For example, \(~\displaystyle\frac{dy}{dx}=\sin y\)
Using the given computer-generated direction field, sketch, by hand, an approximate solution curve that passes through each of the indicated points:
\[ \begin{aligned} \frac{dy}{dx} &= x^2 -y^2 \\ \\ (a) &\;\; y(-2)=1 \\ (b) &\;\; y(-1)=-1 \\ (c) &\;\; y(0)=2 \\ (d) &\;\; y(0)=0 \end{aligned}\]
import numpy as np
import matplotlib.pyplot as plt
plt.style.use('ggplot')
x = np.linspace(-2, 2, 20)
y = np.linspace(-2, 2, 20)
X, Y = np.meshgrid(x, y)
dy = (X*X -Y*Y)
dx = np.ones(dy.shape)
plt.figure(figsize=(5, 5))
plt.quiver(X, Y, dx, dy, color='red')
plt.plot([-2, -1, 0, 0], [1, -1, 2, 0], 'o', color="blue")
plt.xlim([-2.2, 2.2])
plt.ylim([-2.3, 2.2])
plt.xticks(np.arange(-2, 2.5, 0.5))
plt.yticks(np.arange(-2, 2.5, 0.5))
plt.xlabel('x')
plt.ylabel('y')
plt.show()
Isoclines \(y=\pm x\) (slope \(0\)): solutions rise where \(|y|<|x|\) and fall where \(|y|>|x|\).
(a) rises, has a maximum on \(y=-x\) and a minimum on \(y=x\), then rises again \(\quad\) (b) \((-1,-1)\) is a local maximum; to the right the curve falls to \(-\infty\)
(c) falls with slope \(-4\), has a minimum on \(y=x\), then rises \(\quad\) (d) the origin is an inflection point, \(y\approx x^3/3\)
Autonomous first-order DEs, \(~\displaystyle\color{red}{\frac{dy}{dx}=f(y)}\)
An ODE in which the independent variable does not appear explicitly
\[\begin{aligned} \frac{dy}{dx} &= 1+y^2 && \mathrm{autonomous} \\ \frac{dy}{dx} &= 0.2\,xy && \mathrm{nonautonomous} \end{aligned}\]
Critical points, \(~f(c)=0\), \(~\) are constant (or equilibrium) solutions of autonomous DEs
A phase portrait (phase line) is made by putting the critical points on a vertical line, with arrows pointing up or down depending on the sign of \(f(y)\) on the intervals between the points
The horizontal lines \(y=c\) through the critical points divide the \(xy\)-plane into sub-regions (horizontal strips) \(R_i\)
Some conclusions can be drawn about nonconstant solution curves to autonomous DEs
\(~\)
Example \(\,\) Phase portrait and solution curves
\[ \displaystyle \frac{dP}{dt} = P(a-bP) \]
\(~\)
Example \(\,\) Consider the autonomous first-order differential equation
\[\frac{dy}{dx}=y^2-y^4\]
and the initial condition \(y(0)=y_0\). Sketch the graph of a typical solution \(y(x)\) when \(y_0\) has the given values
\[ \begin{aligned} &(a) && \phantom{-1 < }\; y_0 < {-1} \\ &(b) && {-1} < y_0 < 0 \\ &(c) && \phantom{-}0 < y_0 < 1 \\ &(d) && \phantom{-}1 < y_0 \\ \end{aligned}\]
Critical points \(y=-1\) (unstable), \(y=0\) (semi-stable), \(y=1\) (asymptotically stable); inflection points at \(y=\pm1/\sqrt2\).
(a) decreases, \(y\to-1\) as \(x\to-\infty\) and \(y\to-\infty\) at a finite \(x\) \(\quad\) (b) increases from \(-1\) to \(0\)
(c) increases from \(0\) to \(1\) \(\quad\) (d) decreases to \(1\) as \(x\to\infty\)
Consider \(~\displaystyle\frac{dy}{dx}=f(x)\)
A first-order DE of the form \(\displaystyle\frac{dy}{dx}=g(x)h(y)\) is said to be separable, or have separable variables
A separable equation can be rewritten in the form
\[ \color{red}{\frac{1}{h(y)}dy=g(x)dx}\]
which is solved by integrating both sides
\(~\)
Example \(\,\) Solve a separable equation \(\displaystyle\frac{dy}{dx}=y^2-9\), \(\;y(0)=0\)
\[ \begin{aligned} \frac{dy}{(y-3)(y+3)} &= dx \\ &\Downarrow \\ \frac{1}{6} \left [ \frac{1}{y-3} -\frac{1}{y+3} \right ] dy &= dx \end{aligned}\]
\[ \begin{aligned} \frac{1}{6} \ln \left | \frac{y-3}{y+3} \right | &= x+c_1\\ &\Downarrow c=\mp e^{6c_1} \\ y &= 3 \frac{1-ce^{6x}}{1+ce^{6x}} \end{aligned}\]
\(~\)
Example \(\,\) Solve the given differential equation by separation of variables
\(\displaystyle \frac{dy}{dx}=\sin 5 x\)
\(\displaystyle dx +e^{3x} dy = 0\)
\(\displaystyle \frac{dy}{dx} = x\sqrt{1 -y^2}\)
(1) \(y=-\tfrac15\cos5x+c\)
(2) \(y=\tfrac13e^{-3x}+c\)
(3) \(y=\sin\!\left(\tfrac12x^2+c\right)\), \(\,\) plus the singular solutions \(y=\pm1\)
\(~\)
import sympy
from sympy import pi, dsolve
sympy.init_printing()
x = sympy.Symbol('x')
y = sympy.Function('y')
eq = y(x).diff(x) -x *sympy.sqrt(1 -y(x) *y(x))
dsolve(eq, hint='separable')\(\displaystyle y{\left(x \right)} = \sin{\left(C_{1} + \frac{x^{2}}{2} \right)}\)
\(~\)
Example \(\,\) Find an implicit and an explicit solution of the given initial-value problem
\(\displaystyle x^2 \frac{dy}{dx} = y -xy, \;\;y(-1)=-1\)
\(\displaystyle \frac{dx}{dt}=4(x^2 + 1), \;\;x(\pi/4)=1\)
(1) Implicit: \(\ln|y|=-1-\dfrac1x-\ln|x|\); \(\quad\) explicit: \(y=\dfrac1x\,e^{-1-1/x},\ \ x<0\)
(2) Implicit: \(\arctan x=4t-\dfrac{3\pi}4\); \(\quad\) explicit: \(x=\tan\!\left(4t-\dfrac{3\pi}4\right),\ \ \dfrac{\pi}{16}<t<\dfrac{5\pi}{16}\)
\(~\)
t = sympy.Symbol('t')
x = sympy.Function('x')
eq = x(t).diff(t) -4 *(x(t)**2 +1)
dsolve(eq, ics={x(pi/4): 1}, hint='separable')\(\displaystyle x{\left(t \right)} = \tan{\left(4 t + \frac{\pi}{4} \right)}\)
\(~\)
A first-order DE of the form \(\displaystyle a_1(x) \frac{dy}{dx} +a_0(x)y = g(x)~\) is a linear equation in the dependent variable \(y\)
The DE is homogeneous when \(g(x)=0\) ; \(~\)otherwise, \(~\)it is nonhomogeneous
The standard form of a linear DE is obtained by dividing both sides by the lead coefficient
\[\color{red}{\frac{dy}{dx}+P(x)y=f(x)}\]
The standard form equation has the property that its solution \(y\) is the sum of the solution of the associated homogeneous equation \(y_h\) and the particular solution of the nonhomogeneous equation \(y_p\) :
\[\color{red}{y=y_h +y_p}\]
\[ \begin{aligned} \frac{dy_h}{y_h} &= -P(x)dx \\ &\Downarrow \\ \ln |y_h| &= -\int P(x)\,dx +c \,\Rightarrow\, y_h = \bar{c} \exp\left( -\int P(x) \,dx \right) \end{aligned}\]
\[ \begin{aligned} y_h \frac{du}{dx} +& \underbrace{\left (\frac{dy_h}{dx} +P(x) y_h \right )}_{=\,0} u = f(x)\\ du &= \frac{f(x)}{y_h} dx \;\Rightarrow\; u(x) = \displaystyle\int \frac{f(x)}{y_h(x)} dx \\ &\Downarrow \\ y_p &= y_h \displaystyle\int \frac{f(x)}{y_h(x)} dx \end{aligned}\]
\(~\)
Example \(\,\) Find the general solution of the given differential equation:
\(~\displaystyle \frac{dy}{dx} + 2y=0\)
\(~\displaystyle y' +2xy=x^3\)
\(~\displaystyle x\frac{dy}{dx} +2y=3\)
\(~\displaystyle xy' +(1+x)y=e^{-x} \sin 2x\)
(1) \(y=ce^{-2x},\ \ -\infty<x<\infty\)
(2) \(y=\tfrac12x^2-\tfrac12+ce^{-x^2},\ \ -\infty<x<\infty\)
(3) \(y=\dfrac32+\dfrac{c}{x^2},\ \ 0<x<\infty\) \(\,\) (or \(-\infty<x<0\))
(4) \(y=\dfrac{e^{-x}}{x}\left(c-\tfrac12\cos2x\right),\ \ 0<x<\infty\)
\(~\)
Example \(\,\) Solve the given initial-value problem. Give the largest interval \(~I\) over which the solution is defined
\(~\displaystyle y\frac{dx}{dy} -x=2y^2, \;\; y(1)=5\)
\(~\displaystyle (x+1)\frac{dy}{dx}+y = \ln x, \;\;y(1)=10\)
(1) \(x=2y^2-\dfrac{49}{5}y,\ \ 0<y<\infty\)
(2) \(y=\dfrac{x\ln x-x+21}{x+1},\ \ 0<x<\infty\)
\(~\)
x = sympy.Symbol('x')
y = sympy.Function('y')
eq = (x + 1) *y(x).diff(x) +y(x) -sympy.log(x)
dsolve(eq, ics={y(1): 10}, hint='1st_linear')\(\displaystyle y{\left(x \right)} = \frac{x \log{\left(x \right)} - x + 21}{x + 1}\)
\(~\)
Example \(\,\) The given differential equation is not linear in \(y\). \(~\)Nevertheless, find a general solution of the equation
\(~dx=(x+y^2)dy\)
\(~ydx + (2x + xy-3)dy=0\)
(1) \(x=-y^2-2y-2+ce^{y}\)
(2) \(x=\dfrac{3}{y}-\dfrac{3}{y^2}+\dfrac{c}{y^2}e^{-y}\)
\(~\)
Example \(\,\) The sine integral function is defined as
\[ \mathrm{Si}(x)=\int_0^x \frac{\sin t }{t} \,dt\]
where the integrand is defined to be 1 at \(x=0\). Express the solution of the initial value problem
\[x^3 \frac{dy}{dx} + 2x^2 y = 10 \sin x, \;\; y(1)=0\]
in terms of \(\mathrm{Si}(x)\)
\(y=\dfrac{10}{x^2}\left[\mathrm{Si}(x)-\mathrm{Si}(1)\right],\ \ 0<x<\infty\)
\(~\)
A differential expression \(~M(x,y)dx + N(x,y)dy~\) is an exact differential in a region \(R\) of the \(xy\)-plane if it corresponds to the differential of some function \(f(x,y)\):
\[ \color{red}{df(x,y)=\frac{\partial f}{\partial x} dx +\frac{\partial f}{\partial y} dy}\]
Criterion for an exact differential: \(~\)if \(M\), \(N\) and their first partial derivatives are continuous in a rectangular region \(R\), \(~\)then \(M\,dx+N\,dy\) is exact in \(R\) if and only if
\[ \frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}\]
\(M(x,y)\,dx + N(x,y)\, dy=0 ~\) is an exact equation if the left side is an exact differential
\(~\)
Example \(\,\) Solving an exact DE, \(\;2xy\,dx+(x^2-1)\,dy=0\)
\(x^2y-y=c\), \(\,\) i.e. \(\,y=\dfrac{c}{x^2-1}\ \ (x\ne\pm1)\)
\(~\)
Integrating Factor of the first-order linear DE
\[ \begin{aligned} \frac{dy}{dx} +P(x)y &= f(x)\\ &\Downarrow \\ \left ( P(x)y -f(x) \right )dx +dy &= 0\\ &\Downarrow \times \; I(x): \text{ Integrating Factor}\\ I(x)\left ( P(x)y -f(x) \right )dx +I(x)dy &= 0 \\ \\ {\small \text{To be an exact equation }} &\big\Downarrow \\ \frac{\partial}{\partial y} \left\{I(x)\left( P(x){\color{blue}{y}} -f(x) \right) \right \} &= I(x) P(x) =\frac{d}{d x} I(x) \\ &\big\Downarrow \;\; {\color{red}{I(x) = \exp\left(\int P(x) dx\right)}} \end{aligned}\]
Then
\[ \begin{aligned} I(x) \frac{dy}{dx} +I(x) P(x)y &= I(x)f(x) \; \Rightarrow \; \frac{d} {dx}\left\{ I(x)y \right \} = I(x)f(x) \\ &\Downarrow \\ \color{red}{y(x) = I(x)^{-1}y(x_0)I(x_0)} &\color{red}{\,+\,I(x)^{-1} \int_{x_0}^x I(t) f(t)\, dt} \end{aligned}\]
Example \(\,\) Solve \(\displaystyle\frac{dy}{dx} -2xy = 2, \;y(0)=1\)
\[ \begin{aligned} \frac{dy}{dx} -2xy &= 2\\ &\Downarrow \times \;e^{-x^2} \\ \frac{d}{dx}[e^{-x^2}y] &= 2e^{-x^2}\\ y &= c e^{x^2} +2e^{x^2} \int_0^x e^{-t^2} dt\\ & \big\Downarrow \;{\small y(0) = 1 \rightarrow c=1} \\ y &= e^{x^2} \left[ 1 +\sqrt{\pi} \underbrace{\left(\frac{2}{\sqrt{\pi}} \int_0^x e^{-t^2} dt \right)}_{\mathrm{erf}(x)} \right ] \\ &= e^{x^2} \left[1 +\sqrt{\pi} \,\mathrm{erf} (x) \right] \end{aligned}\]
Example \(\,\) Determine whether the given differential equation is exact. If it is exact, solve it
\(~(2x - 1)dx + (3y+7)dy=0\)
\(~(5x + 4y)dx + (4x-8y^2)dy=0\)
\(~(2xy^2-3)dx +(2x^2y+4)dy=0\)
\(~(x^2 -y^2)dx+(x^2-2xy)dy=0\)
(1) Exact: \(x^2-x+\tfrac32y^2+7y=c\)
(2) Exact: \(\tfrac52x^2+4xy-\tfrac83y^3=c\)
(3) Exact: \(x^2y^2-3x+4y=c\)
(4) Not exact (\(M_y=-2y\ne N_x=2x-2y\)); \(\,\) it can be solved by the substitution \(y=ux\) of Section 2.5
\(~\)
x = sympy.Symbol('x')
y = sympy.Function('y')
eq = (2 *x -1) +(3 *y(x) +7) *y(x).diff(x)
dsolve(eq, hint='1st_exact')\(\displaystyle \left[ y{\left(x \right)} = - \frac{\sqrt{C_{1} - 6 x^{2} + 6 x}}{3} - \frac{7}{3}, \ y{\left(x \right)} = \frac{\sqrt{C_{1} - 6 x^{2} + 6 x}}{3} - \frac{7}{3}\right]\)
\(~\)
Example \(\,\) Solve the given initial-value problem
\(~(x+y)^2 dx + (2xy +x^2-1)dy = 0, \;\;y(1)=1\)
\(~(4y + 2t -5)dt + (6y +4t-1)dy=0, \;\;y(-1)=2\)
(1) \(\tfrac13x^3+x^2y+xy^2-y=\tfrac43\)
(2) \(4ty+t^2-5t+3y^2-y=8\)
\(~\)
Example \(\,\) Solve the given differential equation by finding an appropriate integrating factor
\(e^x(xy+y^2)=c\)
Substitution is often used to get a DE in a form that a known procedure can be used to find a solution
Reduction to separation of variables can be facilitated in the DE
\[\frac{dy}{dx}=f(Ax+By+C)~\]
by substituting \(\,u=Ax+By+C, \;B \neq 0\)
\(~\)
Example \(\,\) Solve the IVP \(~\displaystyle\frac{dy}{dx} = (-2x +y)^2 -7, \;y(0)=0\)
Let \(\,u=-2x+y\), then \(~\displaystyle\frac{du}{dx}=-2 + \frac{dy}{dx}~\) giving \(~\displaystyle\frac{du}{dx} = u^2 -9\)
\(y=2x-3\tanh3x\)
\(~\)
x = sympy.Symbol('x')
y = sympy.Function('y')
eq = y(x).diff(x) -(-2 *x +y(x))**2 +7
dsolve(eq, ics={y(0): 0})\(\displaystyle y{\left(x \right)} = \frac{- 2 x e^{6 x} - 2 x + 3 e^{6 x} - 3}{- e^{6 x} - 1}\)
\(~\)
Homogeneous first-order DE
A first-order ordinary DE in the form:
\[\frac{dy}{dx}= - \frac{M(x,y)}{N(x,y)}\]
is of homogeneous type if both functions \(M(x,y)\) and \(N(x,y)\) are homogeneous functions of the same degree \(n\):
\[M(\lambda x, \lambda y) = \lambda^n M(x, y), \;\; N(\lambda x, \lambda y) = \lambda^n N(x, y)\]
Thus, we can let \(\color{blue}{t=1/x}~\) to simplify this quotient to a function \(f\) of the single variable \(y/x\):
\[\frac{M(x, y)}{N(x, y)}=\frac{M(t x,t y)}{N(t x, t y)} = \frac{M\left(1, \tfrac{y}{x}\right)}{N\left(1,\tfrac{y}{x}\right)} =-f\left(\frac{y}{x}\right) = -\frac{dy}{dx}\]
The change of variables \(y=ux\) transforms the original differential equation into the separable form:
\[\frac{du}{f(u)-u}=\frac{dx}{x}\]
\(~\)
Example \(\,\) Solve \((x^2 +y^2) dx +(x^2 -xy) dy = 0\)
\((x+y)^2=Cx\,e^{y/x}\)
\(~\)
Bernoulli DE:
\[~y'+P(x)y =Q(x)y^{n} \;\text{ where }\; n \neq 0~ \text{ and } n \neq 1\]
Bernoulli equations are special because they are nonlinear differential equations with known exact solutions
The substitution \(u=y^{1-n}\) reduces any Bernoulli equation to a linear differential equation
\(~\)
Example \(\,\) Solve \(\displaystyle y'-\frac{2}{x}y=-x^2y^2\)
Changing variables \(\displaystyle u=\tfrac{1}{y}\), \(~\)\(\displaystyle u'=-\tfrac{1}{y^2}y'\) gives the equation
\[ u'+\frac{2}{x}u=x^2\]
\(y=\dfrac{5x^2}{x^5+C}\), \(\,\) plus \(y=0\) (lost when substituting \(u=1/y\))
\(~\)
x = sympy.Symbol('x')
y = sympy.Function('y')
eq = y(x).diff(x) -2 /x *y(x) +x**2 *y(x)**2
dsolve(eq, hint='Bernoulli')\(\displaystyle y{\left(x \right)} = \frac{5 x^{2}}{C_{1} + x^{5}}\)
\(~\)
Riccati DE
Riccati equation is any first-order ordinary differential equation that is quadratic in the unknown function. In other words, it is an equation of the form
\[y'=q_0(x)+q_1(x)y+q_2(x)y^2\]
where \(q_0(x)\neq 0\) and \(q_2(x)\neq 0\). If \(q_0(x)=0\), it is a Bernoulli equation, and if \(q_2(x)=0\), it is a linear equation
The new variable \(v=yq_2\) satisfies an equation
\[v'=v^2+R(x)v+S(x)\]
where \(S(x)=q_2 q_0\) and \(\displaystyle R(x)=q_{1}+\left({\tfrac{q_{2}'}{q_{2}}}\right)\)
Substituting \(\displaystyle v=-\tfrac{u'}{u\,}\), \(~u\) satisfies the linear 2nd order ODE
\[u''-R(x)u'+S(x)u=0\] A solution of this equation will lead to a solution \(\displaystyle y=-\tfrac{u'}{q_2u}\) of the original Riccati equation
\(~\)
Example \(\,\) Solve the given differential equation by using an appropriate substitution
\(~(x-y)dx+xdy=0\)
\(~xdx+(y-2x)dy=0\)
\(~\displaystyle \frac{dy}{dx}=\frac{y-x}{y+x}\)
\(~\displaystyle x\frac{dy}{dx}=y+\sqrt{x^2-y^2}, \;\;x>0\)
(1) \(y=x(c-\ln|x|)\)
(2) \(\ln|y-x|+\dfrac{x}{y-x}=c\), \(\,\) plus \(y=x\)
(3) \(\tfrac12\ln(x^2+y^2)+\arctan\dfrac yx=c\)
(4) \(y=x\sin(\ln x+c)\), \(\,\) plus the singular solutions \(y=\pm x\)
\(~\)
x = sympy.Symbol('x')
y = sympy.Function('y')
eq = x *y(x).diff(x) -y(x) -sympy.sqrt(x**2 -y(x)**2)
dsolve(eq, hint='1st_homogeneous_coeff_best')\(\displaystyle y{\left(x \right)} = - x \sin{\left(C_{1} - \log{\left(x \right)} \right)}\)
\(~\)
Example \(\,\) Solve the given initial-value problem
\(~\displaystyle xy^2 \frac{dy}{dx}=y^3-x^3, \;\;y(1)=2\)
\(~\displaystyle (x^2 +2y^2) \frac{dx}{dy}=xy, \;\; y(-1)=1\)
(1) \(y=x\,(8-3\ln x)^{1/3}\)
(2) \(x^2+y^2=2x^4\); \(\,\) explicitly \(y=-x\sqrt{2x^2-1},\ \ x<-\tfrac1{\sqrt2}\)
\(~\)
Example \(\,\) Solve the given differential equation by using an appropriate substitution
\(~\displaystyle x\frac{dy}{dx} +y =\frac{1}{y^2}\)
\(~\displaystyle 3(1+t^2)\frac{dy}{dt}=2ty(y^3-1)\)
\(~\displaystyle \frac{dy}{dx}=\cos (x+y), \;\;y(0)=\frac{\pi}{4}\)
(1) \(y^3=1+\dfrac{c}{x^3}\)
(2) \(y^{-3}=1+c(1+t^2)\), \(\,\) plus \(y=0\)
(3) \(y=2\arctan\!\left(x+\sqrt2-1\right)-x\)
\(~\)
Numerical method: \(\,\) an algorithm for approximating the unknown solution to a DE
Example \(\,\) Consider \(y'=0.1\sqrt{y}+0.4x^2, \;y(2)=4~\) and approximate \(y(2.5)\) using \(h=0.1\)
Substituting into the general formula for Euler’s method gives
\[y_{n+1}=y_n +h\left(0.1\sqrt{y_n} + 0.4x_n^2 \right)\]
Considering the initial condition and \(n=0\)
\[\scriptsize \begin{aligned} y_1 &= 4+0.1\left( 0.1\sqrt{4} + 0.4 \cdot2^2\right ) =4.1800 \\ y_2 &= 4.18 +0.1\left( 0.1\sqrt{4.18} + 0.4 \cdot2.1^2\right ) =4.3768 \\ &\, \vdots \\ y_5 &= 5.0768 \end{aligned}\]
x0 = 2.0
y0 = 4.0
xf = 2.5
h = 0.1
n = int((xf -x0) /h) + 1
x = np.linspace(x0, xf, n)
y = np.zeros(n)
y[0] = y0
for i in range(1, n):
y[i] = y[i -1] +h *(0.1 *np.sqrt(y[i -1]) +0.4 *x[i -1]**2)
print(f'x = {x[i]: 3.1f}, y = {y[i]: 5.4f}')
plt.figure(figsize=(4, 4))
plt.plot(x, y, 'o')
plt.xlabel("$x$")
plt.ylabel("$y$")
plt.title("$h=0.1$")
plt.show()x = 2.1, y = 4.1800
x = 2.2, y = 4.3768
x = 2.3, y = 4.5914
x = 2.4, y = 4.8244
x = 2.5, y = 5.0768
Example \(\,\) Use Euler’s method to obtain a four-decimal approximation of the indicated value. Carry out the recursion, first using \(h=0.1\) and then using \(h=0.05\)
\[~y'=2x-3y+1, \;\; y(1)=5; \;\;y(1.2)\]
\(h=0.1\): \(\,y(1.2)\approx2.9800\); \(\quad h=0.05\): \(\,y(1.2)\approx3.1151\)
Example \(\,\) Series Circuits - For a series circuit containing a resistor \(R\) and an inductor \(L\),
\(~\)
Kirchhoff’s second law gives
\[L\frac{di}{dt}+Ri=E(t)\]
For a series circuit containing a resistor and a capacitor,
Kirchhoff’s second law gives
\[Ri+\frac{1}{C}q=E(t)\]
where \(\displaystyle i=\frac{dq}{dt}\). \(~\)Then \(~\)\(\displaystyle R\frac{dq}{dt}+\frac{1}{C}q=E(t)\)
\(~\)
Example \(\,\) The population of a community is known to increase at a rate proportional to the number of people present at time \(t\). If an initial population \(P_0\) has doubled in 5 years, how long will it take to triple? To quadruple?
Triples in \(t=5\,\dfrac{\ln3}{\ln2}\approx7.92\) years; \(\,\) quadruples in \(t=5\,\dfrac{\ln4}{\ln2}=10\) years
\(~\)
Example \(\,\) The logistic model
Verhulst proposed a model, called the logistic model, for population growth in 1838. It does not assume unlimited resources. Instead, it assumes there is a carrying capacity \(K\) for the population
This carrying capacity is the stable population level. If the population is above \(K\), then the population will decrease, but if below, then it will increase
For this model, it is assumed that the rate of change \(\frac{dy}{dt}\) of the population \(y\) is proportional to the product of the current population \(y\) and \(K − y\)
\[\frac{dy}{dt}=\alpha y(K-y)\]
\(~\)
Example \(\,\) Radioactive Series:
\[X \xrightarrow{k_1} Y \xrightarrow{k_2} Z\]
The decay of the initial element \(X\) is described by
\[\frac{dx}{dt}=-k_1 x\]
The second element \(Y\) is produced by the decay of \(X\) and is lost through its own spontaneous decay
\[\frac{dy}{dt}=k_1 x -k_2 y\]
The stable element \(Z\) is generated from the decay of \(Y\)
\[\frac{dz}{dt}=k_2 y\]
\(~\)
Example \(\,\) Consider the Lotka-Volterra predator-prey model defined by
\[ \begin{aligned} \frac{dx}{dt} &= -0.1x +0.02xy \\ \frac{dy}{dt} &= 0.2y -0.025xy \end{aligned}\]
where the populations \(x(t)\) (predator) and \(y(t)\) (prey) are measured in the thousands and \(x(0)=6\) and \(y(0)=6\)
# Lotka-Volterra predator-prey model
from scipy import integrate
a, b, c, d = -0.1, 0.02, 0.2, -0.025
def f(t, xy):
x, y = xy
return [a *x +b *x *y, c *y +d *x *y]
# Initial condition and Time span
xy0 = [6, 6]
tf = 400
t_eval = np.linspace(0, tf, 5*tf)
# Numerical Solver
sol = integrate.solve_ivp(f, [0, tf], xy0, t_eval=t_eval)
t = sol.t
xy_t = sol.y.Tfig, ax = plt.subplots(1, 1, figsize=(6, 5))
ax.plot(t, xy_t[:, 0], 'r', label="Predator")
ax.plot(t, xy_t[:, 1], 'b', label="Prey")
ax.set_xlabel("Time")
ax.set_ylabel("Number of animals")
ax.set_xlim(0, tf)
ax.set_ylim(0, 12)
ax.legend()
plt.show()
Both populations are periodic with the same period, \(T\approx45\) (time units)
| Type | Form | Method |
|---|---|---|
| Autonomous | \(y'=f(y)\) | critical points \(f(c)=0\); \(\,\)the sign of \(f\) between them gives stability (phase portrait) |
| Separable | \(y'=g(x)h(y)\) | \(\displaystyle\int\frac{dy}{h(y)}=\int g(x)\,dx\); \(\,\)check the constant solutions \(h(c)=0\) |
| Linear | \(y'+P(x)y=f(x)\) | integrating factor \(I(x)=e^{\int P\,dx}\): \(\ \dfrac{d}{dx}\left[I y\right]=I f\) |
| Exact | \(M\,dx+N\,dy=0\), \(\ M_y=N_x\) | find \(f\) with \(f_x=M\), \(f_y=N\); \(\,\)solution \(f(x,y)=c\) |
| Not exact | \(M_y\ne N_x\) | integrating factor \(\mu(x)=e^{\int (M_y-N_x)/N\,dx}\) \(\,\)if \((M_y-N_x)/N\) depends on \(x\) only; \(\,\)similarly \(\mu(y)=e^{\int (N_x-M_y)/M\,dy}\) |
| Homogeneous | \(M\), \(N\) homogeneous of the same degree | \(y=ux\) \(\,\)(or \(x=vy\)) \(\,\to\,\) separable |
| Bernoulli | \(y'+P(x)y=f(x)y^n\) | \(u=y^{1-n}\) \(\,\to\,\) linear |
| \(y'=f(Ax+By+C)\) | \(u=Ax+By+C\) \(\,\to\,\) separable | |
| Euler’s method | \(y'=f(x,y)\), \(\ y(x_0)=y_0\) | \(y_{n+1}=y_n+h\,f(x_n,y_n)\) |
\(~\)
Solve the given differential equation by separation of variables
2.2: 1. \(\,\displaystyle \frac{dy}{dx}=\sin 5 x\)
Solution
\[ \begin{aligned} \frac{dy}{dx} &=\sin 5x\\ &\Downarrow \\ dy &= \sin 5x \,dx \\ &\Downarrow \\ y &=-\frac{1}{5} \cos 5x +c \end{aligned}\]
2.2: 3. \(\,\displaystyle dx +e^{3x} dy = 0\)
Solution
\[ \begin{aligned} dy &=-e^{-3x} \,dx\\ &\Downarrow \\ y &= \frac{1}{3}e^{-3x} +c \end{aligned}\]
2.2: 21. \(\,\displaystyle \frac{dy}{dx} = x\sqrt{1 -y^2}\)
Solution
\[ \begin{aligned} \frac{1}{\sqrt{1 -y^2}}dy &= x \,dx\\ &\Downarrow \\ \arcsin y &= \frac{1}{2}x^2 +c \\ &\Downarrow \\ y &= \sin \left(\frac{1}{2}x^2 +c \right) \end{aligned}\]
In dividing by \(\sqrt{1-y^2}\) we assumed \(|y|<1\); \(~\)the constant functions \(y=1\) and \(y=-1\) are also (singular) solutions
2.2: 25. Find an implicit or an explicit solution of the given initial-value problem
\[ x^2 \frac{dy}{dx} = y -xy, \;\;y(-1)=-1\]
Solution
\[ \begin{aligned} \frac{1}{y}dy &= \frac{1-x}{x^2} \,dx\\ &\Downarrow \\ \ln |y| &= -\left(\frac{1}{x} +\ln |x| \right) +c \\ &\Downarrow {\small y(-1)=-1 \; \Rightarrow \; c=-1}\\ \ln |y| &= -\left(1 +\frac{1}{x} \right) -\ln |x| \\ &\Downarrow \\ y &= \frac{1}{x} e^{-\left( 1 +\frac{1}{x}\right)} \end{aligned}\]
2.7: 41. \(\,\) Evaporating Raindrop \(\,\) As a raindrop falls, it evaporates while retaining its spherical shape. If we make the further assumptions that the rate at which the raindrop evaporates is proportional to its surface area and that air resistance is negligible, then a model for the velocity \(v(t)\) of the raindrop is
\[ \frac{dv}{dt} + \frac{3(k/\rho)}{(k/\rho) t +r_0} v = g \]
Here \(\rho\) is the density of water, \(r_0\) is the radius of the raindrop at \(t=0\), \(k < 0\) is the constant of proportionality \(\displaystyle \frac{dm}{dt}=4\pi r^2 k\), and the downward direction is taken to be the positive direction
(a)\(\,\) Solve for \(v(t)\) if the raindrop falls from the rest
(b)\(\,\) Show that the radius of the raindrop at time \(t\) is \(r(t)=(k/\rho)t +r_0\)
(c)\(\,\) If \(r_0=0.01\) ft and \(r =0.007\) ft at time \(t=10\) sec after the raindrop falls from a cloud, determine the time at which the raindrop has evaporated completely
Solution
(a)\(\,\)
\[ \begin{aligned} \frac{dv}{dt} &+ \frac{3(k/\rho)}{(k/\rho) t +r_0} v = g \\ &\Downarrow {\small \text{ multiply by the integrating factor }} \\ \frac{d}{dt} \left[ \left(\frac{k}{\rho}t +r_0 \right)^3 v \right] &= \left(\frac{k}{\rho}t +r_0 \right)^3 g\\ &\Downarrow \\ v(t)=\frac{g\rho}{4k} \left( \frac{k}{\rho} t +r_0 \right) &+c\left(\frac{k}{\rho}t +r_0 \right)^{-3} \\ &\Downarrow {\small \text{ } v(0)=0 \; \rightarrow \;c = -\frac{g\rho}{4k} r_0^4} \\ v(t) = \frac{g\rho}{4k} \left(\frac{k}{\rho}t +r_0 \right) & \left[ 1 - \left( \frac{r_0}{ \frac{k}{\rho}t +r_0 } \right)^4 \right] \\ \end{aligned}\]
(b)\(\,\)
\[ \begin{aligned} m &= \frac{4}{3}\pi r^3 \rho\\ &\Downarrow \\ \frac{dm}{dt} &= 4\pi r^2 \rho \frac{dr}{dt}\\ &\Downarrow \text{ } {\scriptstyle \frac{dm}{dt}=4\pi r^2 k}\\ \frac{dr}{dt} &= \frac{k}{\rho} \\ &\Downarrow \text{ } {\scriptstyle r=r_0 \text{ at } t=0} \\ r(t) &= \frac{k}{\rho} t +r_0 \end{aligned}\]
(c)\(\,\)
\[ \begin{aligned} r(10) &= \frac{k}{\rho} \times 10 +0.01 =0.007 \; \Rightarrow \; \frac{k}{\rho} = -0.0003\\ &\Downarrow \\ r(t) &= -0.0003 t + 0.01 = 0 \; \Rightarrow \; t =\frac{0.01}{0.0003} \approx 33.3 \;\text{sec} \end{aligned}\]
\(~\)
1. \(~\)Find an explicit solution of the given initial value problem
\[\sqrt{1-y^2} \,dx -\sqrt{1-x^2}\,dy=0, \;y(0)=\sqrt{3}/2\]
Solution
Step 1: \(~\)Rearrange the Equation
We write the equation in differential form:
\[\sqrt{1 - y^2} \, dx = \sqrt{1 - x^2} \, dy\]
Separate variables:
\[\frac{dx}{\sqrt{1 - x^2}} = \frac{dy}{\sqrt{1 - y^2}}\]
Step 2: \(~\)Integrate Both Sides
\[\int \frac{dx}{\sqrt{1 - x^2}} = \int \frac{dy}{\sqrt{1 - y^2}}\]
These are standard integrals:
\(\displaystyle\scriptsize \int \frac{dx}{\sqrt{1 - x^2}} = \arcsin x + C\)
\(\displaystyle\scriptsize \int \frac{dy}{\sqrt{1 - y^2}} = \arcsin y + C\)
So:
\[\arcsin x = \arcsin y + C\]
Step 3: \(~\)Solve for the Constant Using the Initial Condition
We’re given \(y(0) = \frac{\sqrt{3}}{2}\), so plug in \(x = 0\), \(y = \frac{\sqrt{3}}{2}\):
\[\arcsin(0) = \arcsin\left(\frac{\sqrt{3}}{2}\right) + C\]
\[0 = \frac{\pi}{3} + C \Rightarrow C = -\frac{\pi}{3}\]
Step 4: \(~\)Write the Explicit Solution
From earlier:
\[\arcsin x = \arcsin y - \frac{\pi}{3}\]
Solve for \(y\):
\[\arcsin y = \arcsin x + \frac{\pi}{3}\]
Now apply \(\sin\) to both sides:
\[y = \sin\left(\arcsin x + \frac{\pi}{3}\right)\]
or, expanding the sine, \(~\displaystyle y=\frac{x}{2}+\frac{\sqrt{3}}{2}\sqrt{1-x^2}\)
This is a solution only where \(\cos\left(\arcsin x+\frac{\pi}{3}\right)\ge 0\), \(~\)because \(\frac{dy}{dx}=\frac{\cos\left(\arcsin x+\pi/3\right)}{\sqrt{1-x^2}}\) must equal \(\frac{\sqrt{1-y^2}}{\sqrt{1-x^2}}\ge 0\). \(~\)Hence \(\arcsin x+\frac{\pi}{3}\le\frac{\pi}{2}\), \(~\)i.e. the solution is defined for \(-1<x\le\frac{1}{2}\)
\(~\)
2. \(~\)Find an explicit solution of the given initial value problem
\[\frac{dy}{dx} =-y\ln y, \;y(0)=e\]
Solution
Step 1: \(~\)Separate Variables
We want to isolate \(y\) and \(x\):
\[\frac{dy}{dx} = -y \ln y \quad \Rightarrow \quad \frac{1}{y \ln y} \, dy = -dx\]
Step 2: \(~\)Integrate Both Sides
Let’s handle the left-hand side:
Let \(u = \ln y\), then:
\[\frac{du}{dy} = \frac{1}{y} \Rightarrow du = \frac{1}{y} \, dy\]
So:
\[\frac{1}{y \ln y} \, dy = \frac{1}{u} \cdot du = \frac{du}{u}\]
Thus, we have:
\[\int \frac{dy}{y \ln y} = \int \frac{du}{u} = \ln |\ln y|\]
So now integrate both sides:
\[\int \frac{dy}{y \ln y} = \int -dx \quad \Rightarrow \quad \ln|\ln y| = -x + C\]
Step 3: \(~\)Solve for \(y(x)\)
Start with:
\[\ln|\ln y| = -x + C\]
Exponentiate both sides:
\[|\ln y| = e^{-x + C} = A e^{-x}, \quad A > 0\]
So:
\[\ln y = \pm A e^{-x}\]
We drop the \(\pm\) by using a general constant \(B = \pm A\), so:
\[\ln y = B e^{-x} \Rightarrow y = e^{B e^{-x}}\]
Step 4: \(~\)Apply Initial Condition \(y(0) = e\)
Plug in:
\[y(0) = e^{B e^0} = e^{B} = e \Rightarrow B = 1\]
Final Answer:
\[y(x) = e^{e^{-x}}\]
This is the explicit solution satisfying the differential equation and the initial condition \(y(0) = e\)
\(~\)
3. \(~\)Solve the given initial-value problem
\[\frac{dy}{dx} = \frac{3x+2y}{3x+2y+2}, \;y(-1)=-1\]
Solution
Step 1: \(~\)Make a Substitution
Let us simplify the expression by setting:
\[u = 3x + 2y\]
Differentiate both sides with respect to x:
\[\frac{du}{dx} = 3 + 2\frac{dy}{dx}\]
Now solve for \(\frac{dy}{dx}\) from this:
\[\frac{dy}{dx} = \frac{1}{2} \left( \frac{du}{dx} - 3 \right)\]
But also, from the original equation:
\[\frac{dy}{dx} = \frac{u}{u + 2}\]
So equating both expressions:
\[\frac{1}{2} \left( \frac{du}{dx} - 3 \right) = \frac{u}{u + 2}\]
So:
\[\frac{du}{dx} = 3 + \frac{2u}{u + 2}\]
Step 2: \(~\)Simplify the Right-Hand Side
We simplify and so:
\[\frac{du}{dx} = \frac{5u + 6}{u + 2}\]
Step 3: \(~\)Separate Variables and Integrate
We separate:
\[\frac{u + 2}{5u + 6} \, du = dx\]
Let’s integrate the left-hand side. Let’s simplify:
\[\int \frac{u + 2}{5u + 6} \, du = \int \left[ \frac{1}{5} + \frac{4}{5(5u + 6)} \right] du = \frac{1}{5} u + \frac{4}{25} \ln |5u + 6| + C\]
So:
\[x = \frac{1}{5} u + \frac{4}{25} \ln |5u + 6| + C\]
Step 4: \(~\)Substitute Back \(u = 3x + 2y\)
Recall \(u = 3x + 2y\). So we substitute:
\[x = \frac{1}{5}(3x + 2y) + \frac{4}{25} \ln |5(3x + 2y) + 6| + C\]
This is an implicit solution
Step 5: \(~\)Apply the Initial Condition \(y(-1) = -1\)
Recall \(u = 3x + 2y\), so at \(x = -1\), \(y = -1\):
\[u = 3(-1) + 2(-1) = -5 \Rightarrow 5u + 6 = -25 + 6 = -19\]
Then:
\[-1 = \frac{1}{5}(-5) + \frac{4}{25} \ln(19) + C \Rightarrow C = - \frac{4}{25} \ln 19\]
Final Answer:
\[ x = \frac{1}{5}(3x + 2y) + \frac{4}{25} \ln |5(3x + 2y) + 6| - \frac{4}{25} \ln 19 \]
\(~\)
4. \(~\)Solve the given differential equation
\[\frac{dy}{dx} = \tan^2(x+y)\]
Solution
Step 1: \(~\)Make a Substitution
Let’s simplify the expression using substitution.
Let:
\[u = x + y \quad \Rightarrow \quad \frac{du}{dx} = 1 + \frac{dy}{dx}\]
From the original equation:
\[\frac{dy}{dx} = \tan^2(x + y) = \tan^2 u\]
So:
\[\frac{du}{dx} = 1 + \tan^2 u = \sec^2 u\]
Step 2: \(~\)Separate Variables and Integrate
\[\frac{du}{\sec^2 u} = dx \quad \Rightarrow \quad \cos^2 u \, du = dx\]
So integrate both sides:
\[\int \cos^2 u \, du = \int \frac{1 + \cos(2u)}{2} \, du = \frac{1}{2} \left[ u + \frac{1}{2} \sin(2u) \right]\]
So:
\[\frac{1}{2} u + \frac{1}{4} \sin(2u) = x + C\]
Step 3: \(~\)Substitute Back \(u = x + y\)
\[\frac{1}{2}(x + y) + \frac{1}{4} \sin(2(x + y)) = x + C\]
where \(C\) is a constant of integration. This is an implicit solution to the original differential equation
\(~\)
5. \(~\) Find the solution of period \(2\pi\) of the equation
\[ y' +(\cos x) y = \sin 2x\]
Solution
Step 1: \(~\)Solve Using the Integrating Factor Method
Integrating Factor
\[\mu(x) = e^{\int \cos x \, dx} = e^{\sin x}\]
Multiply both sides of the ODE by \(\mu(x) = e^{\sin x}\):
\[e^{\sin x} y’ + e^{\sin x} \cos x \, y = e^{\sin x} \sin 2x\]
Left-hand side becomes:
\[\frac{d}{dx}\left( e^{\sin x} y \right) = e^{\sin x} \sin 2x\]
Step 2: \(~\)Integrate Both Sides
With \(s=\sin x\), \(~ds=\cos x\,dx\) and \(\sin 2x=2\sin x\cos x\), the integral is elementary:
\[\int e^{\sin x} \sin 2x \, dx = \int 2s\,e^{s}\,ds = 2(s-1)e^{s} = 2(\sin x-1)\,e^{\sin x}\]
So:
\[y(x) = e^{-\sin x} \left( 2(\sin x-1)e^{\sin x} + C \right) = 2(\sin x - 1) + C e^{-\sin x}\]
Step 3: \(~\)Impose Periodicity
Both \(2(\sin x-1)\) and \(e^{-\sin x}\) have period \(2\pi\), \(~\)so every solution has period \(2\pi\), \(~\)whatever the value of \(C\):
\[\color{red}{y(x) = 2(\sin x - 1) + C e^{-\sin x}}\]
(The periodicity can also be seen without evaluating the integral: \(~e^{-\sin x}\) is \(2\pi\)-periodic and \(\int_0^{2\pi} e^{\sin t}\sin 2t\,dt = 0\), \(~\)because the integrand changes sign under \(t\mapsto \pi-t\).) \(~\)Check: \(~y=2\sin x-2\) gives \(y'+(\cos x)y = 2\cos x+2\sin x\cos x-2\cos x=\sin 2x\)
\(~\)
6. \(~\) Find the solution of period \(2\pi\) of the equation
\[ y' +3 y = \cos x\]
Solution
Step 1: \(~\)General Solution via Integrating Factor
This is a first-order linear ODE. Let’s solve it using the integrating factor method.
Let:
\(\mu(x) = e^{\int 3 \, dx} = e^{3x}\)
Multiply both sides of the equation by \(e^{3x}\):
\[e^{3x} y’ + 3 e^{3x} y = e^{3x} \cos x \Rightarrow \frac{d}{dx}(e^{3x} y) = e^{3x} \cos x\]
Now integrate both sides:
\[e^{3x} y = \int e^{3x} \cos x \, dx\]
Step 2: \(~\)Compute the Integral
\[\int e^{3x} \cos x \, dx = \frac{e^{3x}}{10}(3 \cos x + \sin x) + C\]
Thus:
\[e^{3x} y = \frac{e^{3x}}{10}(3 \cos x + \sin x) + C \Rightarrow y(x) = \frac{1}{10}(3 \cos x + \sin x) + C e^{-3x}\]
Step 3: \(~\)Impose Periodicity
We are asked to find a solution with period \(2\pi\). Let’s analyze:
So to obtain a periodic solution, we must take: \(C = 0\)
Final Answer:
\[y(x) = \frac{1}{10}(3 \cos x + \sin x)\]
\(~\)
7. \(~\) Solve the given differential equation by using an appropriate substitution
\[ \frac{dy}{dx} = \sin(x+y)\]
Solution
Step 1: \(~\)Use Substitution
Let:
\[u = x + y \quad \Rightarrow \quad \frac{du}{dx} = 1 + \frac{dy}{dx}\]
Substitute into the derivative of \(u\): \[\frac{du}{dx} = 1 + \sin u\]
Step 2: \(~\)Separate Variables
We now have the separable equation:
\[\frac{du}{1 + \sin u} = dx\]
We simplify the left-hand side:
\[\frac{1}{1 + \sin u} = \frac{1 - \sin u}{(1 + \sin u)(1 - \sin u)} = \frac{1 - \sin u}{1 - \sin^2 u} = \frac{1 - \sin u}{\cos^2 u}\]
Now integrate:
\[\int \sec^2 u \, du = \tan u, \quad \int \sec u \tan u \, du = \sec u\]
So:
\[\int \frac{du}{1 + \sin u} = \tan u - \sec u \;\Rightarrow\; \tan u - \sec u = x + C\]
Step 3: \(~\)Substitute Back \(u = x + y\)
\[\tan(x + y) - \sec(x + y) = x + C\]
This is the implicit general solution to the differential equation. \(~\)In separating variables we divided by \(1+\sin u\), \(~\)so the constant solutions \(\sin u=-1\), \(~\)i.e. \(y=-x-\frac{\pi}{2}+2k\pi\), \(~\)must be added
\(~\)
8. \(~\) Find a general solution of the equation
\[ydx+(2x+xy-3)dy=0\]
Solution
\[\begin{aligned} ydx +(2x+xy-3)&dy =0 \\ &\Downarrow \\ \frac{dx}{dy} + \left(1 + \frac{2}{y} \right)x &= \frac{3}{y} \\ &\Downarrow \\ \frac{d}{dy} \left[ x \,y^2 e^y \right ] &= 3y e^y \\ &\Downarrow \\ \color{red}{x = \frac{3}{y}\left(1-\frac{1}{y} \right)}\, & \color{red}{\, + \,Cy^{-2} e^{-y}} \end{aligned}\]
\[~\text{or}~\]
\[\begin{aligned} ydx \;+\;&(2x+xy-3)dy =0 \\ &\Downarrow \;\; \times \, y e^y\\ y^2 e^y dx \;+\;&y e^y (2x+xy-3)dy =0 \\ &\Downarrow \\ \frac{\partial f}{\partial x} = y^2 e^y \;\;&\rightarrow \;\; f(x,y) =y^2 e^y x + h(y)\\ &\Downarrow \\ \frac{\partial f}{\partial y} = ye^y(2x +xy) +\frac{dh}{dy} \;\;&\rightarrow \;\; \frac{dh}{dy} = -3y e^y\\ &\Downarrow \\ h(y) =-3e^y (y-1) \;\;&\rightarrow \;\; f(x,y) =y^2 e^y x -3e^y (y-1) = C \\ &\Downarrow \\ \color{red}{x = \frac{3}{y}\left(1-\frac{1}{y} \right)}\, & \color{red}{\, + \,Cy^{-2} e^{-y}} \end{aligned}\]
\(~\)
9. \(~\) Solve the differential equation
\[ xv\frac{dv}{dx} +v^2=32x \]
Solution \[ \begin{aligned} xv\frac{dv}{dx} +v^2 &=32x \\ &\Downarrow \\ \frac{dv}{dx} +\frac{1}{x} v &= \frac{32}{v}\\ &\Downarrow \;\; u = v^2, \; \frac{du}{dx} = 2v \frac{dv}{dx}\\ \frac{du}{dx} +\frac{2}{x}u &= 64\\ &\Downarrow \\ \frac{d}{dx} \left( u x^2 \right ) &= 64x^2 \\ &\Downarrow \\ u = \frac{64}{3} x + \frac{C}{x^2} \;\;&\rightarrow \;\; \color{red}{v = \pm\sqrt{\frac{64}{3} x + \frac{C}{x^2}}} \end{aligned} \]
\(~\)
10. \(~\) Solve the initial-value problem
\[\frac{dy}{dt} + 2(t+1) y^2=0, \;\;y(0)=-\frac{1}{8}\]
and give the largest interval \(I\) on which the solution is defined:
Solution
\[\begin{aligned} \frac{dy}{dt} &+2(t+1)y^2 = 0 \\ &\Downarrow \\ \frac{dy}{y^2} &= -2(t+1) \, dt \\ &\Downarrow \\ y &= \frac{1}{t^2 + 2t + c}, \;\; y(0)=-\frac{1}{8} \\ &\Downarrow \\ \color{red}{y} &\color{red}{= \frac{1}{(t-2)(t+4)}, \;\;-4<t<2} \end{aligned} \]
\(~\)