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Practice
Differential Equations
Differential Equations: Problems with Solutions
By
Prof. Hernando Guzman Jaimes (University of Zulia - Maracaibo, Venezuela)
Problem 1
What is the solution to this differential equation?
$dx+e^{3x}dy=0$
$y=\frac{1}{3}e^{3x}+C$
$y=e^{x}+C$
$y=\frac{1}{3}e^{-3x}+C$
$y=\frac{1}{3e^{-3x}}+C$
Solution:
We apply variable separation method.
$dx+e^{3x}dy=0\Longrightarrow dx=-e^{3x}dy$
Then $e^{-3x}dx=-dy$ and now we can apply integration.
$\int e^{-3x}dx=-\int dy\Longrightarrow -\frac{1}{3}e^{-3x}=-y+C$ and after rewriting this
$y=\frac{1}{3}e^{-3x}+C$
Problem 2
After applying variable separation method to $dy-(y-1)^{2}dx=0$
we get
$-\frac{1}{y-1}=x+C$
$\frac{1}{x+1}=y+C$
$\frac{1}{y+1}=-x+C$
$\frac{1}{y-x}=C$
Solution:
Let we apply variable separation method.
So $dy-(y-1)^{2}dx=0\Longrightarrow dy=(y-1)^{2}dx $
Then $(y-1)^{-2}dy=dx$ and now we can apply integration, then
$\int (y-1)^{-2}dy=\int dx$ in the first integral we can apply a substitution
$u=y-1\Longrightarrow dy=dy$ then $\int u^{-2}du=\int dx\Longrightarrow -\frac{1}{u}=x+C$
Since $u=y-1$ we get $-\frac{1}{y-1}=x+C$
Problem 3
Solve the differential equation using variable separation
$\dfrac{dy}{dx}+2xy^{2}=0$
$y=\frac{1}{x^{2}+C}$
$y=\frac{1}{x^{2}+x+C}$
$y=\left( \frac{1}{x+C}\right) ^{2}$
$y=\frac{x^{2}}{x+C}$
Solution:
$\dfrac{dy}{dx}+2xy^{2}=0\Longrightarrow \dfrac{dy}{dx}=-2xy^{2}$ then $\frac{1}{y^{2}}dy=-2xdx$
and we apply integration $\int \frac{1}{y^{2}}dy=-2\int xdx\Longrightarrow -\frac{1}{y}=-x^{2}+C$
$\Longrightarrow y=\frac{1}{x^{2}+C}$
Problem 4
Solve: $xy'+y=e^{x}$, where $y(1)=2$
$y=x+C$
$y=\frac{C}{x^{2}}$
$y=Cx$
$y=\frac{C}{x}$
Solution:
We can apply variable separation to
$xy'+y=0$, $y(1)=2$
Then $x\frac{dy}{dx}+y=0\Longrightarrow x\frac{dy}{dx}=-y$
So $\frac{1}{y}dy=-\frac{1}{x}dx$ and after integrating $\int \frac{1}{y}dy=-\int \frac{1}{x}dx\Longrightarrow \ln y+\ln x=C$
We can apply the logarithm property $\ln \left( xy\right) =\ln C\Longrightarrow xy=C$
So $y=\frac{C}{x}$ is the solution.
Problem 5
Solve the differential equation
$\dfrac{dy}{dx}=e^{3x+2y}\qquad y(0)=1$
$2e^{3x}=\frac{3}{e^{2}}+2$
$3e^{-2y}=\frac{3}{e^{2}}+2$
$2e^{3x}+3e^{-2y}=0$
$2e^{3x}+3e^{-2y}=\frac{3}{e^{2}}+2$
Solution:
$\dfrac{dy}{dx}=e^{3x+2y}$
If we apply the exponential properties, we get
$\dfrac{dy}{dx}=e^{3x+2y}=e^{3x}e^{2y}$
We need to apply variable separation
$e^{-2y}dy=e^{3x}dx$ and integrate $\int e^{-2y}dy=\int e^{3x}dx$
$-\frac{1}{2}e^{-2y}=\frac{1}{3}e^{3x}+C\Longrightarrow \frac{1}{3}e^{3x}+\frac{1}{2}e^{-2y}=C$
Then $2e^{3x}+3e^{-2y}=C$ (a) we use the initial condition and so, we obtain the value of $C$,
$y(0)=1\Longrightarrow 2e^{0}+3e^{-2}=C=\frac{3}{e^{2}}+2=C$
After substituting in (a) we get
$2e^{3x}+3e^{-2y}=\frac{3}{e^{2}}+2$
Problem 6
$e^{x}y\dfrac{dy}{dx}=e^{-y}+e^{-2x-y}$
After applying variable separation we get:
$e^{y}=-e^{-x}-\frac{1}{3}e^{-3x}+C$
$y=-e^{-x}-\frac{1}{3}e^{-3x}+C$
$e^{y}\left( y-1\right) =-e^{-x}-\frac{1}{3}e^{-3x}+C$
$\left( y-1\right) =\frac{1}{3}e^{-3x}+C$
Solution:
$e^{x}y\dfrac{dy}{dx}=e^{-y}+e^{-2x-y}\Longrightarrow e^{x}y\dfrac{dy}{dx}=e^{-x}(1+e^{-2x})$
We apply variable separation method.
$ye^{y}dy=e^{-x}(1+e^{-2x})dx$
$ye^{y}dy=(e^{-x}+e^{-3x})dx$ so we can integrate
$\int ye^{y}dy=\int (e^{-x}+e^{-3x})dx$ the left integral is by parts method
and right integral is immediate.
Then $\int ye^{y}dy=e^{y}\left( y-1\right)$ (by parts)
and $\int (e^{-x}+e^{-3x})dx=-e^{-x}-\frac{1}{3}e^{-3x}+C$
So
$e^{y}\left( y-1\right) =-e^{-x}-\frac{1}{3}e^{-3x}+C$
Problem 7
Solve the differential equation:
$y\ln x\dfrac{dx}{dy}=(\dfrac{y+1}{x})^{2}$
$y(1)=1$
$\frac{1}{9}x^{3}=\frac{y^{2}}{2}+2y+\ln y-\frac{47}{18}$
$\frac{1}{3}x^{3}\ln x=\frac{y^{2}}{2}+2y+\ln y-\frac{47}{18}$
$\frac{1}{3}x^{3}\ln x-\frac{1}{9}x^{3}=\frac{y^{2}}{2}$
$\frac{1}{3}x^{3}\ln x-\frac{1}{9}x^{3}=\frac{y^{2}}{2}+2y+\ln y-\frac{47}{18}$
Solution:
We apply variable separation method.
$y\ln x\dfrac{dx}{dy}=(\dfrac{y+1}{x})^{2}$
$x^{2}\ln xdx=\frac{1}{y}(y+1)^{2}dy\Longrightarrow x^{2}\ln xdx=\left( y+2+\frac{1}{y}\right) dy$
Then we integrate, so, $\int x^{2}\ln xdx=\int \left( y+2+\frac{1}{y}\right) dy$
the left integral is solved using integration by parts.
$\int x^{2}\ln xdx=\allowbreak \frac{1}{3}x^{3}\ln x-\frac{1}{9}x^{3}$ (by parts) and $\int \left( y+2+\frac{1}{y}\right) dy=\frac{y^{2}}{2}+2y+\ln y+C$
The solution is $\frac{1}{3}x^{3}\ln x-\frac{1}{9}x^{3}=\frac{y^{2}}{2}+2y+\ln y+C,$
now we obtain the value of $C$
Consider the initial condiction $y(1)=1$, $y=1$ when $x=1$
$\frac{1}{3}1^{3}\ln 1-\frac{1}{9}1^{3}=\frac{1^{2}}{2}+2(1)+\ln 1+C\Longrightarrow -\frac{1}{9}=\frac{1}{2}+2+C$
So $C=-\frac{1}{9}-\frac{1}{2}-2=-\frac{47}{18}$ and replacing in the general solution we get $\frac{1}{3}x^{3}\ln x-\frac{1}{9}x^{3}=\frac{y^{2}}{2}+2y+\ln y-\frac{47}{18}$
Problem 8
$\sqrt{1-y^{2}}dx-\sqrt{1-x^{2}}dy=0,$
$y(0)=\dfrac{\sqrt{3}}{2}$
$\arcsin \frac{x}{y}=\frac{\pi }{3}$
$\arcsin x=\arcsin y-\frac{\pi }{3}$
$\arcsin x=\arcsin y+\frac{2\pi }{3}$
$\arcsin x=\arcsin y-\dfrac{\sqrt{3}}{2}$
Solution:
$\sqrt{1-y^{2}}dx-\sqrt{1-x^{2}}dy=0$
Apply variable separation method.
$\frac{1}{\sqrt{1-x^{2}}}dx=\frac{1}{\sqrt{1-y^{2}}}dy$
Consider the integraion formula
$\int \frac{du}{\sqrt{a-u^{2}}}=\arcsin \frac{u}{a}+C$ so $\arcsin x=\arcsin y+C$ is the solution.
We use the initial condition, $\arcsin 0=\arcsin \dfrac{\sqrt{3}}{2}+C$
Then $0=\frac{1}{3}\pi +C\Longrightarrow C=-\frac{\pi }{3}$ and replacing
$\int \frac{1}{\sqrt{1-x^{2}}}dx=\int \frac{1}{\sqrt{1-y^{2}}}dy \Longrightarrow \arcsin x=\arcsin y-\frac{\pi }{3}$
Problem 9
The result of applying variable separation method in
$\csc y$ $dx+\sec ^{2}x$ $dy=0$ is:
$\frac{1}{2}\left[ x+\frac{1}{2}\sin 2x\right] =\cos y+C$
$\frac{1}{2}\left[ x+\frac{1}{2}\cos 2x\right] =\sin y+C$
$\frac{1}{2}\left[ 2x+\frac{1}{2}\sin x\right] =\cos 2y+C$
$\frac{1}{2}\left[ 2x+2\sin 2x\right] =\cos 2y+C$
Solution:
$\csc y dx+\sec^{2}x$ $dy=0$
Step 1: Apply variable separation.
$\csc y$ $dx=-\sec ^{2}x$ $dy\Longrightarrow \cos^{2}xdx=-\sin ydy$
Step 2: Apply trigonometric identity.
$\frac{1}{2}\left( 1+cos2x\right) dx=-\sin ydy$
Step 3: Integrate, $\frac{1}{2}\int \left( 1+cos2x\right) dx=-\int \sin ydy$
Result:
$\frac{1}{2}\left[ x+\frac{1}{2}\sin 2x\right] =\cos y+C$
Problem 10
$\sin 3x$ $dx+2y$ $\cos ^{3}3x$ $dy=0$
$y(\pi )=0$
$\frac{1}{6}\sec 3x=-y+\frac{1}{6}$
$\frac{1}{6}\sec ^{3}3x=-y^{3}+\frac{1}{6}$
$\frac{1}{6}\sec ^{2}3x=-y^{2}+\frac{1}{6}$
$\frac{1}{6}\csc ^{2}3x=y^{2}+\frac{1}{6}$
Solution:
$\sin 3x$ $dx+2y$ $\cos ^{3}3x$ $dy=0$
$\Longrightarrow \sin 3x\ dx=-2y\ dy$
Apply variable seperation $\frac{\sin 3x}{\cos^{3}3x} dx=-2ydy$
$\int \frac{\sin 3x}{\cos^{3}3x} dx=-2\int ydy$
In the left integral, we apply the substitution $u=\cos 3x\Longrightarrow du=-3\sin 3xdx$
So
$-\frac{1}{3}\int u^{-3}du=-y^{2}+C\Longrightarrow \frac{1}{6}u^{-2}=-y^{2}+C $
and since
$u=\cos 3x$ then $\frac{1}{6\cos^{2}3x}=-y^{2}+C$ is the general solution It is given that $y(\pi )=0$, so
$\frac{1}{6\cos^{2}\left( 3\pi \right) }=0+C\Longrightarrow \frac{1}{6}=C$
The solution is
$\frac{1}{6}\sec^{2}3x=-y^{2}+\frac{1}{6}$
Problem 11
Find the general solution to the differential equation
$(e^{y}+1)^{2}e^{-y}dx+(e^{x}+1)^{3}e^{-x}dy=0$
$\frac{1}{e^{y}+1}+\frac{1}{2\left( e^{x}+1\right) ^{2}}=C$
$e^{y}+1+2\left( e^{x}+1\right) ^{2}=C$
$\frac{1}{e^{x}+1}+\frac{1}{2\left( e^{y}+1\right) ^{2}}=C$
$e^{x}+1+2\left( e^{y}+1\right) ^{2}=C$
Solution:
After applying variable separation, we get
$(e^{y}+1)^{2}e^{-y}dx=-(e^{x}+1)^{3}e^{-x}dy\Longrightarrow \frac{e^{x}}{(e^{x}+1)^{3}}dx=-\frac{e^{y}}{(e^{y}+1)^{2}}dy$
Then $\int \frac{e^{x}}{(e^{x}+1)^{3}}dx=-\int \frac{e^{y}}{(e^{y}+1)^{2}}dy$
We can solve both integrals using the substitution
$u=e^{x}+1$
$du=e^{x}dx$ and
$v=e^{y}+1$
$dv=e^{y}dy$ so,
$\int u^{-3}du=-\int v^{-2}dv$
$-\frac{1}{2}u^{-2}=\frac{1}{v}+C$ and since
$u=e^{x}+1,v=e^{y}+1$
We get
$-\frac{1}{2\left( e^{x}+1\right) ^{2}}=\frac{1}{e^{y}+1}+C\Longrightarrow \frac{1}{e^{y}+1}+\frac{1}{2\left( e^{x}+1\right) ^{2}}=C$
Problem 12
$x(1+y^{2})^{1/2}dx=y(1+x^{2})^{1/2}dy$
$\sqrt{1+y^{2}}=\sqrt{1-x^{2}}+C$
$\sqrt{1+x^{2}}=\sqrt{1+y^{2}}+C$
$1+x^{2}=\sqrt{1+y^{2}}+C$
$\sqrt{1+x^{2}}=1+y^{2}+C$
Solution:
We apply variable separation $x(1+x^{2})^{-1/2}dx=y(1+y^{2})^{-1/2}dy$
and integrate $\int x(1+x^{2})^{-1/2}dx=\int y(1+y^{2})^{-1/2}dy$
We can solve both integrals using the substitution
$u=1+x^{2}$
$du=2xdx$ and $v=1+y^{2}$
$dv=2ydy$ so, we get
$\frac{1}{2}\int (u)^{-1/2}du=\frac{1}{2}\int (v)^{-1/2}dv\Longrightarrow 2.\frac{1}{2}\sqrt{u}=2.\frac{1}{2}\sqrt{v}+C$
So
$\sqrt{1+x^{2}}=\sqrt{1+y^{2}}+C$ is the solution.
Problem 13
What is the solution to the differential equation?
$\dfrac{dy}{dx}+y=e^{3x}$
$y=e^{x}\left[ \frac{1}{4}e^{-4x}\right] =\frac{1}{4}e^{3x}$
$y=e^{x}\left[ \frac{1}{4}e^{4x}\right] =\frac{1}{4}e^{-3x}$
$y=e^{-x}\left[ \frac{1}{4}e^{-4x}\right] =\frac{1}{4}e^{-3x}$
$y=e^{-x}\left[ \frac{1}{4}e^{4x}\right] =\frac{1}{4}e^{3x}$
Solution:
It is a linear differential equation in the form
$\dfrac{dy}{dx}+P(x)y=Q(x)$
Apply the formula $y=e^{-\int P(x)dx}\left[ \int Q(x)e^{\int P(x)dx}dx\right] $
where $\dfrac{dy}{dx}+y=e^{3x}$, $P(x)=1$ and $Q(x)=e^{3x}$
$\int P(x)dx=\int dx=x$ and $\int Q(x)e^{\int P(x)dx}dx=\int e^{3x}e^{x}dx=\int e^{4x}dx=\frac{1}{4}e^{4x}$
The solution is $y=e^{-x}\left[ \frac{1}{4}e^{4x}\right] =\frac{1}{4}e^{3x}$
Problem 14
After applying the linear equation method to $y'+3x^{2}y=x^{2}$ we get
$y=e^{x^{3}}\left[ \frac{1}{3}e^{-x^{3}}\right] =\frac{1}{3}$
$y=e^{-x^{3}}\left[ \frac{1}{4}e^{-x^{3}}\right] =\frac{1}{4}$
$y=e^{-x^{3}}\left[ \frac{1}{3}e^{x^{3}}\right] =\frac{1}{3}$
$y=e^{x^{3}}\left[ \frac{1}{4}e^{x^{3}}\right] =\frac{1}{4}$
Solution:
Apply the formula $y=e^{-\int P(x)dx}\left[ \int Q(x)e^{\int P(x)dx}dx\right] $
$y'+3x^{2}y=x^{2}\Longrightarrow P(x)=3x^{2}\qquad Q(x)=x^{2}$ then
$\int P(x)dx=\int 3x^{2}dx=x^{3}$ and $\int Q(x)e^{\int P(x)dx}dx=\int x^{2}e^{x^{3}}dx=\frac{1}{3}\int e^{u}du=\frac{1}{3}e^{u}=\frac{1}{3}e^{x^{3}}$
The solution is $y=e^{-x^{3}}\left[\frac{1}{3}e^{x^{3}}\right] =\frac{1}{3}$
$y=\frac{1}{3}$
Problem 15
Solve the differential equation
$y'+2xy=x^{3}$
$y=e^{-x^{2}y}\left[ \frac{1}{y^{2}}e^{x^{2}y}\left( x^{2}y-1\right) \right] $
$y=e^{x^{2}y}\left[ \frac{1}{y^{2}}e^{-x^{2}y}\left( x^{2}y-1\right) \right] $
$y=e^{x^{2}y}\left[ \frac{1}{y^{2}}e^{-x^{2}y}\left( x^{2}y+1\right) \right] $
$y=e^{-x^{2}y}\left[ \frac{1}{x^{2}}e^{x^{2}y}\left( xy^{2}-1\right) \right] $
Solution:
First apply the formula: $y=e^{-\int P(x)dx}\left[ \int Q(x)e^{\int P(x)dx}dx\right] $
$y'+2xy=x^{3}$,
$P(x)=2xy\qquad Q(x)=x^{3}$
Then
$\int P(x)dx=\int 2xydx=x^{2}y$ and $\int Q(x)e^{\int P(x)dx}dx=\int x^{3}e^{x^{2}y}dx$
To solve this integral we substitute, $u=x^{2}\Longrightarrow du=2xdx\Longrightarrow \frac{1}{2}du=xdx$
$\int x^{3}e^{x^{2}y}dx=\frac{1}{2}\int ue^{yu}du$ $=$ $\frac{1}{y^{2}}e^{uy}\left( uy-1\right) $
and after integrating by parts we get
$\int x^{3}e^{x^{2}y}dx=\frac{1}{y^{2}}e^{x^{2}y}\left( x^{2}y-1\right)$
The solution to this differential equation is
$y=e^{-x^{2}y}\left[ \frac{1}{y^{2}}e^{x^{2}y}\left( x^{2}y-1\right) \right]$
Problem 16
This differential equation is a linear function
$x^{2}y'+xy=1$
What is the solution?
$y=x\ln x+C$
$y=\frac{1}{x}\ln x+C$
$y=x^{2}\ln x+C$
$y=x+\ln x+C$
Solution:
$y=e^{-\int P(x)dx}\left[ \int Q(x)e^{\int P(x)dx}dx\right] $
$x^{2}y'+xy=1\Longrightarrow y'+\frac{1}{x}y=\frac{1}{x^{2}}$
Then $P(x)=\frac{1}{x}$ and $Q(x)=\frac{1}{x^{2}}$
$\int P(x)dx=\int \frac{1}{x}dx=\ln x$ and $\int \frac{1}{x^{2}}e^{\ln x}dx=\int \frac{1}{x}dx=\ln x$
So the solution is $y=e^{-\ln x}\left[ \ln x\right] =\frac{1}{e^{\ln x}}\ln x=\frac{1}{x}\ln x$
$y=\frac{1}{x}\ln x+C$
Problem 17
What is the solution to this linear equation?
$y'=2y+x^{2}+5$
$\mathbf{y}=4\left( 2x^{2}-2x+11\right) $
$\mathbf{y}=4\left( 2x^{2}+2x-11\right) $
$\mathbf{y}=\frac{1}{4}\left( 2x^{2}-2x-11\right) $
$\mathbf{y}=\frac{1}{4}\left( 2x^{2}+2x+11\right) $
Solution:
Apply the formula $y=e^{-\int P(x)dx}\left[\int Q(x)e^{\int P(x)dx}dx\right]$
$y'=2y+x^{2}+5\Longrightarrow y'-2y=x^{2}+5\Longrightarrow P(x)=-2\qquad Q(x)=x^{2}+5$
and $\int P(x)dx=-2x$ so $\int Q(x)e^{\int P(x)dx}dx=\int (x^{2}+5)e^{-2x}dx$
The last integral must be solved using itegration by parts, so
$\int (x^{2}+5)e^{-2x}dx= -\frac{1}{4}e^{-2x}\left(2x^{2}+2x+11\right)$
Then the solution to the differential equation
is $y=e^{2x}\left[ -\frac{1}{4}e^{-2x}\left( 2x^{2}+2x+11\right) \right] =\frac{1}{4}\left(2x^{2}+2x+11\right)$
Problem 18
Solve this linear differential equation
$x\dfrac{dy}{dx}-y=x^{2}\sin (x)$
$y=-y\cos x+c$
$y=-x\cos y+C$
$y=x\cos y+C$
$y=-x\cos x+C$
Solution:
$x\dfrac{dy}{dx}-y=x^{2}\sin (x)\Longrightarrow \dfrac{dy}{dx}-\frac{1}{x}y=x\sin (x)$
Then $P(x)=-\frac{1}{x}$
$Q(x)=x\sin (x)$ and we apply $y=e^{-\int P(x)dx}\left[ \int Q(x)e^{\int P(x)dx}dx\right] $
So $\int P(x)dx=\int -\frac{1}{x}dx=-\ln x$ and $\int Q(x)e^{\int P(x)dx}dx=\int x\sin (x)e^{-\ln x}dx=\int \sin xdx=-\cos x+C$
Then the solution to the differential equation is
$y=e^{\ln x}\left[ -\cos x\right] =-x\cos x+C$
Problem 19
Solve this linear differential equation.
$(1+x)\dfrac{dy}{dx}-xy=x+x^{2}$
$y=\frac{1}{x+1}\left( x^{2}-3x+3\right) $
$y=-\frac{1}{x+1}\left( x^{2}+3x+3\right) $
$y=\frac{1}{x+1}\left( x^{2}-3x-3\right) $
$y=\left( x+1\right) \left( x^{2}+3x+3\right) $
Solution:
$(1+x)\dfrac{dy}{dx}-xy=x+x^{2}\Longrightarrow \dfrac{dy}{dx}-\frac{x}{(1+x)}y=\frac{x(1+x)}{(1+x)}=x$
Then $P(x)=-\frac{x}{(1+x)}\qquad Q(x)=x$
$y=e^{-\int P(x)dx}\left[ \int Q(x)e^{\int P(x)dx}dx\right]$ so $\int P(x)dx=-\int \frac{x}{(1+x)}dx$
Apply sustitution method to evaluate the integral $-\int \frac{x}{(1+x)}dx= \ln \left( x+1\right) -x.$
Now $\int xe^{\left( \ln \left(x+1\right) -x\right) }dx=\int x(x+1)e^{-x}dx$
After applying integration by parts we get
$\int x(x+1)e^{-x}dx=\allowbreak -e^{-x}\left( x^{2}+3x+3\right) $
Apply the formula
$y=e^{-\int P(x)dx}\left[ \int Q(x)e^{\int P(x)dx}dx\right] =e^{x-\ln (x+1)}\left[ -e^{-x}\left( x^{2}+3x+3\right) \right] $
Then $y=-e^{x-\ln (x+1)-x}\left( x^{2}+3x+3\right) =-\frac{1}{x+1}\left(x^{2}+3x+3\right)$
is the solution to the differential equation.
Problem 20
Solve this linear differential equation
$x^{2}y'+x(x+2)y=e^{x}$
$\mathbf{y}=\frac{1}{2}\left( x^{2}e^{x}\right) $
$\mathbf{y}=\frac{1}{2}\left( \frac{1}{x^{2}}e^{x}\right) $
$\mathbf{y}=\frac{1}{2}\left( \frac{1}{x^{2}}e^{x^{2}}\right) $
$\mathbf{y}=\frac{1}{2}\left( e^{x}\right) $
Solution:
$x^{2}y'+x(x+2)y=e^{x}$
Then $y'+\frac{1}{x}(x+2)y=\frac{1}{x^{2}}e^{x}$ so
$P(x)=\frac{1}{x}(x+2)=1+\frac{2}{x}\qquad Q(x)=\frac{1}{x^{2}}e^{x}$ and we apply
$y=e^{-\int P(x)dx}\left[ \int Q(x)e^{\int P(x)dx}dx\right]$ so $\int P(x)dx=\int (1+\frac{2}{x})dx=x+2\ln x=x+\ln x^{2}$ and $\int Q(x)e^{\int P(x)dx}dx=\int \frac{1}{x^{2}}e^{x}e^{x+\ln x^{2}}dx=\int e^{2x}dx=\frac{1}{2}e^{2x}$
The solution to the differential equation is
$y=e^{-\left( x+\ln x^{2}\right) }\left[ \frac{1}{2}e^{2x}\right] =\frac{1}{2}e^{2x}(e^{-x}\frac{1}{x^{2}})=\frac{1}{2}\left( \frac{1}{x^{2}}e^{x}\right)$
Problem 21
What is the solution to the differential equation $(x+1)\dfrac{dy}{dx}+y=\ln x$ with initial value $y(1)=10$?
$y=\frac{1}{x+1}\left[ x\left( \ln x-1\right) \right] +\frac{21}{2}$
$y=x+1\left[ x\left( \ln x-1\right) \right] +\frac{21}{2}$
$y=\frac{1}{x+1}\left[ x\left( \ln x+1\right) \right] +\frac{21}{2}$
$y=x\left[ \left( x+1\right) \left( \ln x-1\right) \right] +\frac{21}{2}$
Solution:
$(x+1)\dfrac{dy}{dx}+y$ $=\ln x\Longrightarrow \dfrac{dy}{dx}+\frac{1}{(x+1)}y=\frac{1}{(x+1)}\ln x$
Here $P(x)=\frac{1}{(x+1)},\qquad Q(x)=\frac{1}{(x+1)}\ln x$ so we apply
$y=e^{-\int P(x)dx}\left[ \int Q(x)e^{\int P(x)dx}dx\right]$
Since $\int P(x)dx=\int \frac{1}{(x+1)}dx$
$\int P(x)dx=\ln (x+1)$ and $\int Q(x)e^{\int P(x)dx}dx=\int \frac{1}{(x+1)}\ln xe^{\ln (x+1)}dx=\int \ln xdx= x\left( \ln x-1\right) $
After replacing we get
$y=e^{-\ln (x+1)}\left[ x\left( \ln x-1\right) \right] =\frac{1}{x+1}\left[ x\left( \ln x-1\right) \right] +C$ is the solution
Now we use the initial condicion $y(1)=10$ so
$10=\frac{1}{1+1}\left[ 1\left( \ln 1-1\right) \right] +C=-\frac{1}{2}+C\Longrightarrow C=\frac{21}{2}$
The particular solution is $y=\frac{1}{x+1}\left[ x\left( \ln x-1\right) \right] +\frac{21}{2}$
Problem 22
What is the solution with initial value $y(0)=-1$?
$y'+(\tan x)y=\cos ^{2}x$
$y=\cos x-\sin x$
$y=\cos x+\sin x$
$y=\cos x\cdot\sin x-1$
$y=\cos x\cdot\sin x$
Solution:
$y'+(\tan x)y=\cos ^{2}x\Longrightarrow P(x)=\tan x,\qquad Q(x)=\cos^{2}x$ so we apply
$y=e^{-\int P(x)dx}\left[ \int Q(x)e^{\int P(x)dx}dx\right]$ since $\int P(x)dx=\int \tan xdx=\allowbreak -\ln \left( \cos x\right)=\ln (\sec x)$ and $\int Q(x)e^{\int P(x)dx}dx=\int \cos ^{2}xe^{\ln (\sec x)}dx=\int \cos ^{2}x\sec xdx=\int \cos xdx=\sin x$
If we apply the formula, we get
$y=e^{-\ln (\sec x)}\left[ \sin x\right] =\cos x \cdot \sin x+C$ and considering that
$y(0)=-1\Longrightarrow -1=\cos 0 \cdot \sin 0+C\Longrightarrow C=-1$
The particular solution is $y=\cos x \cdot \sin x-1$
Problem 23
Find the particular solution to the differential equation $\dfrac{dy}{dx}+2xy=f(x),y(0)=2$ where
$f(x)=\left\{ \begin{array}{c} x,\text{ \ }0\leq x<1 \\ 0,\text{ \ \ \ \ \ \ \ \ }x\geq 1 \end{array} \right\}$
$y=e^{x^{2}}\left[ \frac{1}{2}\left( e-6\right) \right] +\left( \frac{5}{2}+\frac{1}{2}e\right) $
$y=e^{-x^{2}}\left[ \frac{1}{2}\left( e+1\right) \right] +\left( \frac{5}{2}+\frac{1}{2}e\right) $
$y=e^{x^{2}}\left[ \frac{1}{2}\left( e-1\right) \right] +\left( \frac{5}{2}-e\right) $
$y=e^{-x^{2}}\left[ \frac{1}{2}\left( e-1\right) \right] +\left( \frac{5}{2}-\frac{1}{2}e\right) $
Solution:
$P(x)=2x\qquad Q(x)=f(x)$ and $\int P(x)dx=x^{2}$
and $\int Q(x)e^{\int P(x)dx}dx=\overset{1}{\underset{0}{\int }}xe^{x^{2}}dx=\frac{1}{2} \overset{1}{\underset{0}{\int }}e^{u}du=\frac{1}{2}\left( e-1\right) $
So the general solution is $y=e^{-\int P(x)dx}\left[ \int Q(x)e^{\int P(x)dx}dx\right] $
$y=e^{-x^{2}}\left[ \frac{1}{2}\left( e-1\right) \right] +C$ now, if we use the initial condition
$y(0)=2$ we get the particular solution
$2=e^{-0}\left[ \frac{1}{2}\left( e-1\right) \right] +C\Longrightarrow C=2-\frac{1}{2}e+\frac{1}{2}=\frac{5}{2}-\frac{1}{2}e$
The particular solution is $y=e^{-x^{2}}\left[ \frac{1}{2}\left(e-1\right) \right] +\left( \frac{5}{2}-\frac{1}{2}e\right) $
Problem 24
Find the particular solution to the differential equation $(1+x^{2})\frac{dy}{dx}+2xy=f(x),y(0)=0,$ where
$f(x)=\left\{ \begin{array}{c} x,\text{ \ }0\leq x<1 \\ 0,\text{ \ \ \ \ \ \ \ \ }x\geq 1 \end{array} \right\} $
$y=\frac{1}{2\left( 1+x^{2}\right) }+C$
$y=\frac{\left( 1+x^{2}\right) }{2}+C$
$y=\left( 1+x^{2}\right) +C$
$y=\frac{1}{\left( 1-x^{2}\right) }+C$
Solution:
$(1+x^{2})\frac{dy}{dx}+2xy=f(x)\Longrightarrow \frac{dy}{dx}+\frac{2x}{(1+x^{2})}y=\frac{1}{(1+x^{2})}f(x)$
here $P(x)=\frac{2x}{(1+x^{2})}\qquad Q(x)=\frac{1}{(1+x^{2})}f(x)$ then
$\int P(x)dx=\int \frac{2x}{(1+x^{2})}dx=\ln (1+x^{2})$ and $\int Q(x)e^{\int P(x)dx}dx=\int \frac{1}{(1+x^{2})}f(x)e^{\ln (1+x^{2})}dx=\int f(x)dx=\underset{0}{ \overset{1}{\int }}xdx=\frac{1}{2}\left[ 1-0\right] =\frac{1}{2}$
We apply the formula
$y=e^{-\int P(x)dx}\left[ \int Q(x)e^{\int P(x)dx}dx\right] =y=e^{-\ln(1+x^{2})}\left[ \frac{1}{2}\right] =\frac{1}{2\left( 1+x^{2}\right) }+C$
So the general solution is $y=\frac{1}{2\left( 1+x^{2}\right) }+C$
We can find a particular solution $y(0)=0$ then
$0=\frac{1}{2\left( 1+0^{2}\right) }+C\Longrightarrow C=-\frac{1}{2}$
So $y=\frac{1}{2\left( 1+x^{2}\right) }+C$ is the particular solution.
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