Differential Equations: Problems with Solutions

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

$y=\frac{1}{3}e^{-3x}+C$

Solution:
We apply the 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

$u=y-1\Longrightarrow du=dy$ then $\int u^{-2}du=\int dx\Longrightarrow -\frac{1}{u}=x+C$

Since $u=y-1$ we get $-\frac{1}{y-1}=x+C$

Solution:
$\dfrac{dy}{dx}+2xy^{2}=0\Longrightarrow \dfrac{dy}{dx}=-2xy^{2}$ then $\frac{1}{y^{2}}dy=-2xdx$

and we apply integration $\int \frac{1}{y^{2}}dy=-2\int xdx\Longrightarrow -\frac{1}{y}=-x^{2}+C$
$\Longrightarrow y=\frac{1}{x^{2}+C}$

Solution:
Notice that the left-hand side is exactly the derivative of a product:

[tex]xy'+y=\left( xy\right) '[/tex]

So the equation can be written as [tex]\left( xy\right) '=e^{x}[/tex] and we integrate both sides

[tex]xy=\int e^{x}dx=e^{x}+C[/tex]

Then [tex]y=\dfrac{e^{x}+C}{x}[/tex] is the general solution.

Now we use the initial condition [tex]y(1)=2[/tex]

[tex]2=\dfrac{e^{1}+C}{1}\Longrightarrow C=2-e[/tex]

The particular solution is [tex]y=\dfrac{e^{x}+2-e}{x}[/tex]

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
$2e^{3x}+3e^{-2y}=\frac{3}{e^{2}}+2$

Solution:
$e^{x}y\dfrac{dy}{dx}=e^{-y}+e^{-2x-y}\Longrightarrow e^{x}y\dfrac{dy}{dx}=e^{-y}(1+e^{-2x})$
We apply the variable separation method, moving every [tex]y[/tex] to the left and every [tex]x[/tex] to the right.
$ye^{y}dy=e^{-x}(1+e^{-2x})dx$

$ye^{y}dy=(e^{-x}+e^{-3x})dx$ 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
$e^{y}\left( y-1\right) =-e^{-x}-\frac{1}{3}e^{-3x}+C$

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 $\frac{1}{3}x^{3}\ln x-\frac{1}{9}x^{3}=\frac{y^{2}}{2}+2y+\ln y-\frac{47}{18}$

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 integration formula
$\int \frac{du}{\sqrt{a^{2}-u^{2}}}=\arcsin \frac{u}{a}+C$ 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
$\int \frac{1}{\sqrt{1-x^{2}}}dx=\int \frac{1}{\sqrt{1-y^{2}}}dy \Longrightarrow \arcsin x=\arcsin y-\frac{\pi }{3}$

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:
$\frac{1}{2}\left[ x+\frac{1}{2}\sin 2x\right] =\cos y+C$

Solution:
$\sin 3x$ $dx+2y$ $\cos ^{3}3x$ $dy=0$
Apply variable separation $\frac{\sin 3x}{\cos^{3}3x} dx=-2ydy$
$\int \frac{\sin 3x}{\cos^{3}3x} dx=-2\int ydy$
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
$\frac{1}{6}\sec^{2}3x=-y^{2}+\frac{1}{6}$

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
$-\frac{1}{2\left( e^{x}+1\right) ^{2}}=\frac{1}{e^{y}+1}+C\Longrightarrow \frac{1}{e^{y}+1}+\frac{1}{2\left( e^{x}+1\right) ^{2}}=C$

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.

Solution:
It is a linear differential equation of the form

[tex]\dfrac{dy}{dx}+P(x)y=Q(x)[/tex]

so we apply the formula

[tex]y=e^{-\int P(x)dx}\left[ \int Q(x)e^{\int P(x)dx}dx+C\right][/tex]

where [tex]P(x)=1[/tex] and [tex]Q(x)=e^{3x}[/tex].

[tex]\int P(x)dx=\int dx=x[/tex] and [tex]\int Q(x)e^{\int P(x)dx}dx=\int e^{3x}e^{x}dx=\int e^{4x}dx=\dfrac{1}{4}e^{4x}[/tex]

The general solution is

[tex]y=e^{-x}\left[ \dfrac{1}{4}e^{4x}+C\right] =\dfrac{1}{4}e^{3x}+Ce^{-x}[/tex]

The term [tex]Ce^{-x}[/tex] is the solution of the associated homogeneous equation [tex]y'+y=0[/tex]; leaving it out gives only one particular solution, not the general one.

Solution:
Apply the formula

[tex]y=e^{-\int P(x)dx}\left[ \int Q(x)e^{\int P(x)dx}dx+C\right][/tex]

[tex]y'+3x^{2}y=x^{2}\Longrightarrow P(x)=3x^{2}\qquad Q(x)=x^{2}[/tex]

then [tex]\int P(x)dx=\int 3x^{2}dx=x^{3}[/tex] and, with the substitution [tex]u=x^{3},\ du=3x^{2}dx[/tex],

[tex]\int Q(x)e^{\int P(x)dx}dx=\int x^{2}e^{x^{3}}dx=\dfrac{1}{3}\int e^{u}du=\dfrac{1}{3}e^{u}=\dfrac{1}{3}e^{x^{3}}[/tex]

The general solution is

[tex]y=e^{-x^{3}}\left[ \dfrac{1}{3}e^{x^{3}}+C\right] =\dfrac{1}{3}+Ce^{-x^{3}}[/tex]

The constant [tex]y=\dfrac{1}{3}[/tex] on its own is only one particular solution; the term [tex]Ce^{-x^{3}}[/tex] comes from the constant of integration and must be kept.

Solution:
First apply the formula: $y=e^{-\int P(x)dx}\left[ \int Q(x)e^{\int P(x)dx}dx\right] $

$y'+2xy=x^{3}$,
$P(x)=2x\qquad Q(x)=x^{3}$

Then
$\int P(x)dx=\int 2xdx=x^{2}$ and $\int Q(x)e^{\int P(x)dx}dx=\int x^{3}e^{x^{2}}dx$

To solve this integral we substitute, $u=x^{2}\Longrightarrow du=2xdx\Longrightarrow \frac{1}{2}du=xdx$

$\int x^{3}e^{x^{2}}dx=\frac{1}{2}\int ue^{u}du$ $=$ $\frac{1}{2}e^{u}\left( u-1\right) $

and after integrating by parts we get

$\int x^{3}e^{x^{2}}dx=\frac{1}{2}e^{x^{2}}\left( x^{2}-1\right)+C$

Substitute back into the formula for y:
$y=e^{−x^2}\left[\frac{1}{2}​e^{x^2}(x^2−1)+C\right]$

Simplify:
$y=\frac12(x^2−1)+Ce^{−x^2}$

Solution:
Divide by [tex]x^{2}[/tex] to bring the equation to standard form

[tex]x^{2}y'+xy=1\Longrightarrow y'+\dfrac{1}{x}y=\dfrac{1}{x^{2}}[/tex]

Then [tex]P(x)=\dfrac{1}{x}[/tex] and [tex]Q(x)=\dfrac{1}{x^{2}}[/tex], and we apply

[tex]y=e^{-\int P(x)dx}\left[ \int Q(x)e^{\int P(x)dx}dx+C\right][/tex]

[tex]\int P(x)dx=\int \dfrac{1}{x}dx=\ln x[/tex] and [tex]\int \dfrac{1}{x^{2}}e^{\ln x}dx=\int \dfrac{1}{x}dx=\ln x[/tex]

So the general solution is

[tex]y=e^{-\ln x}\left[ \ln x+C\right] =\dfrac{1}{x}\left( \ln x+C\right) =\dfrac{\ln x+C}{x}[/tex]

The constant belongs inside the brackets and is therefore divided by [tex]x[/tex] as well.

A quick check: [tex]y=\dfrac{\ln x+C}{x}[/tex] gives [tex]xy=\ln x+C[/tex], so [tex]\left( xy\right) '=\dfrac{1}{x}[/tex], and multiplying by [tex]x[/tex] returns [tex]x^{2}y'+xy=1[/tex].

Solution:
Apply the formula [tex]y=e^{-\int P(x)dx}\left[\int Q(x)e^{\int P(x)dx}dx+C\right][/tex]

[tex]y'=2y+x^{2}+5\Longrightarrow y'-2y=x^{2}+5\Longrightarrow P(x)=-2\qquad Q(x)=x^{2}+5[/tex]

and [tex]\int P(x)dx=-2x[/tex] so [tex]\int Q(x)e^{\int P(x)dx}dx=\int \left( x^{2}+5\right) e^{-2x}dx[/tex]

The last integral is solved using integration by parts, so

[tex]\int \left( x^{2}+5\right) e^{-2x}dx=-\dfrac{1}{4}e^{-2x}\left( 2x^{2}+2x+11\right)[/tex]

Then the solution to the differential equation is

[tex]y=e^{2x}\left[ -\dfrac{1}{4}e^{-2x}\left( 2x^{2}+2x+11\right) +C\right] =-\dfrac{1}{4}\left( 2x^{2}+2x+11\right) +Ce^{2x}[/tex]

Note the two details that are easy to lose here: the minus sign in front of [tex]\dfrac{1}{4}[/tex] stays, and the constant [tex]C[/tex] sits inside the brackets, so it is multiplied by [tex]e^{2x}[/tex] as well.

Solution:
[tex]x\dfrac{dy}{dx}-y=x^{2}\sin (x)\Longrightarrow \dfrac{dy}{dx}-\dfrac{1}{x}y=x\sin (x)[/tex]

Then [tex]P(x)=-\dfrac{1}{x}[/tex] and [tex]Q(x)=x\sin (x)[/tex], and we apply

[tex]y=e^{-\int P(x)dx}\left[ \int Q(x)e^{\int P(x)dx}dx+C\right][/tex]

So [tex]\int P(x)dx=\int -\dfrac{1}{x}dx=-\ln x[/tex] and

[tex]\int Q(x)e^{\int P(x)dx}dx=\int x\sin (x)e^{-\ln x}dx=\int \sin x\,dx=-\cos x[/tex]

Then the solution to the differential equation is

[tex]y=e^{\ln x}\left[ -\cos x+C\right] =x\left( C-\cos x\right) =-x\cos x+Cx[/tex]

The constant has to stay inside the brackets, so it is multiplied by [tex]x[/tex] together with the rest. Writing [tex]-x\cos x+C[/tex] instead would not satisfy the equation.

Solution:
[tex](1+x)\dfrac{dy}{dx}-xy=x+x^{2}\Longrightarrow \dfrac{dy}{dx}-\dfrac{x}{1+x}y=\dfrac{x(1+x)}{1+x}=x[/tex]

Then [tex]P(x)=-\dfrac{x}{1+x}\qquad Q(x)=x[/tex] and we apply

[tex]y=e^{-\int P(x)dx}\left[ \int Q(x)e^{\int P(x)dx}dx+C\right][/tex]

[tex]\int P(x)dx=-\int \dfrac{x}{1+x}dx=-\int \left( 1-\dfrac{1}{1+x}\right) dx=\ln \left( x+1\right) -x[/tex]

Now [tex]\int xe^{\ln \left( x+1\right) -x}dx=\int x\left( x+1\right) e^{-x}dx[/tex] and after integration by parts

[tex]\int x\left( x+1\right) e^{-x}dx=-e^{-x}\left( x^{2}+3x+3\right)[/tex]

Since [tex]e^{-\int P(x)dx}=e^{x-\ln \left( x+1\right) }=\dfrac{e^{x}}{x+1}[/tex], the general solution is

[tex]y=\dfrac{e^{x}}{x+1}\left[ -e^{-x}\left( x^{2}+3x+3\right) +C\right] =\dfrac{Ce^{x}-\left( x^{2}+3x+3\right) }{x+1}[/tex]

Taking [tex]C=0[/tex] gives the particular solution [tex]y=-\dfrac{x^{2}+3x+3}{x+1}[/tex], but the term [tex]\dfrac{Ce^{x}}{x+1}[/tex] solves the homogeneous equation and belongs in the general answer.

Solution:
Divide by [tex]x^{2}[/tex]

[tex]x^{2}y'+x(x+2)y=e^{x}\Longrightarrow y'+\dfrac{x+2}{x}y=\dfrac{1}{x^{2}}e^{x}[/tex]

so [tex]P(x)=\dfrac{x+2}{x}=1+\dfrac{2}{x}\qquad Q(x)=\dfrac{1}{x^{2}}e^{x}[/tex] and we apply

[tex]y=e^{-\int P(x)dx}\left[ \int Q(x)e^{\int P(x)dx}dx+C\right][/tex]

[tex]\int P(x)dx=\int \left( 1+\dfrac{2}{x}\right) dx=x+2\ln x=x+\ln x^{2}[/tex]

[tex]\int Q(x)e^{\int P(x)dx}dx=\int \dfrac{1}{x^{2}}e^{x}\,e^{x+\ln x^{2}}dx=\int e^{2x}dx=\dfrac{1}{2}e^{2x}[/tex]

Since [tex]e^{-\left( x+\ln x^{2}\right) }=\dfrac{e^{-x}}{x^{2}}[/tex], the general solution is

[tex]y=\dfrac{e^{-x}}{x^{2}}\left[ \dfrac{1}{2}e^{2x}+C\right] =\dfrac{e^{x}}{2x^{2}}+\dfrac{Ce^{-x}}{x^{2}}[/tex]

The second term is the solution of the homogeneous equation, and the constant is divided by [tex]x^{2}[/tex] because it sits inside the brackets.

Solution:
[tex](x+1)\dfrac{dy}{dx}+y=\ln x\Longrightarrow \dfrac{dy}{dx}+\dfrac{1}{x+1}y=\dfrac{\ln x}{x+1}[/tex]

Here [tex]P(x)=\dfrac{1}{x+1},\qquad Q(x)=\dfrac{\ln x}{x+1}[/tex] so we apply

[tex]y=e^{-\int P(x)dx}\left[ \int Q(x)e^{\int P(x)dx}dx+C\right][/tex]

Since [tex]\int P(x)dx=\int \dfrac{1}{x+1}dx=\ln \left( x+1\right)[/tex] and

[tex]\int Q(x)e^{\int P(x)dx}dx=\int \dfrac{\ln x}{x+1}e^{\ln \left( x+1\right) }dx=\int \ln x\,dx=x\left( \ln x-1\right)[/tex]

After replacing we get the general solution

[tex]y=e^{-\ln \left( x+1\right) }\left[ x\left( \ln x-1\right) +C\right] =\dfrac{x\left( \ln x-1\right) +C}{x+1}[/tex]

The constant stays inside the brackets, so it is divided by [tex]x+1[/tex] as well. Now we use the initial condition [tex]y(1)=10[/tex]

[tex]10=\dfrac{1\left( \ln 1-1\right) +C}{1+1}=\dfrac{C-1}{2}\Longrightarrow C=21[/tex]

The particular solution is [tex]y=\dfrac{x\left( \ln x-1\right) +21}{x+1}[/tex]

A quick check: at [tex]x=1[/tex] this gives [tex]\dfrac{-1+21}{2}=10[/tex], as required.

Solution:
[tex]y'+(\tan x)y=\cos ^{2}x\Longrightarrow P(x)=\tan x,\qquad Q(x)=\cos ^{2}x[/tex] so we apply

[tex]y=e^{-\int P(x)dx}\left[ \int Q(x)e^{\int P(x)dx}dx+C\right][/tex]

since [tex]\int P(x)dx=\int \tan x\,dx=-\ln \left( \cos x\right) =\ln \left( \sec x\right)[/tex]

and [tex]\int Q(x)e^{\int P(x)dx}dx=\int \cos ^{2}x\,e^{\ln \left( \sec x\right) }dx=\int \cos ^{2}x\sec x\,dx=\int \cos x\,dx=\sin x[/tex]

If we apply the formula, we get

[tex]y=e^{-\ln \left( \sec x\right) }\left[ \sin x+C\right] =\cos x\left( \sin x+C\right) =\sin x\cos x+C\cos x[/tex]

The constant belongs inside the brackets, so it is multiplied by [tex]\cos x[/tex] as well - this is where a stray [tex]+C[/tex] on its own would give a function that does not satisfy the equation.

Now we use the initial condition

[tex]y(0)=-1\Longrightarrow -1=\sin 0\cos 0+C\cos 0=C\Longrightarrow C=-1[/tex]

The particular solution is [tex]y=\sin x\cos x-\cos x=\cos x\left( \sin x-1\right)[/tex]

Solution:
The equation is linear with [tex]P(x)=2x[/tex], so [tex]\int P(x)dx=x^{2}[/tex] and the integrating factor is [tex]e^{x^{2}}[/tex]. Since the right-hand side is given by two different formulas, we solve on each interval separately.

On [tex]0\leq x<1[/tex], we have [tex]f(x)=x[/tex], so

[tex]\int Q(x)e^{\int P(x)dx}dx=\int xe^{x^{2}}dx=\dfrac{1}{2}e^{x^{2}}[/tex]

[tex]y=e^{-x^{2}}\left[ \dfrac{1}{2}e^{x^{2}}+C\right] =\dfrac{1}{2}+Ce^{-x^{2}}[/tex]

The initial condition [tex]y(0)=2[/tex] gives [tex]2=\dfrac{1}{2}+C\Longrightarrow C=\dfrac{3}{2}[/tex], so

[tex]y=\dfrac{1}{2}+\dfrac{3}{2}e^{-x^{2}}[/tex]

On [tex]x\geq 1[/tex], we have [tex]f(x)=0[/tex], so the equation becomes [tex]y'+2xy=0[/tex] and

[tex]y=Ke^{-x^{2}}[/tex]

The constant [tex]K[/tex] is fixed by requiring the solution to be continuous at [tex]x=1[/tex]

[tex]\dfrac{1}{2}+\dfrac{3}{2}e^{-1}=Ke^{-1}\Longrightarrow K=\dfrac{1}{2}e+\dfrac{3}{2}=\dfrac{e+3}{2}[/tex]

So the particular solution is

[tex]y=\left\{ \begin{array}{ll} \dfrac{1}{2}+\dfrac{3}{2}e^{-x^{2}} , & 0\leq x<1 \\ \dfrac{e+3}{2}e^{-x^{2}} , & x\geq 1 \end{array} \right.[/tex]

A warning about a common slip: the integral [tex]\int xe^{x^{2}}dx[/tex] must stay indefinite. Replacing it by the definite integral [tex]\int_{0}^{1}xe^{x^{2}}dx=\dfrac{1}{2}\left( e-1\right)[/tex] turns it into a number, and the function obtained that way does not satisfy the equation.

Solution:
Notice first that the left-hand side is already the derivative of a product:

[tex]\left( 1+x^{2}\right) \dfrac{dy}{dx}+2xy=\dfrac{d}{dx}\left[ \left( 1+x^{2}\right) y\right][/tex]

(this is what the integrating factor [tex]e^{\int \frac{2x}{1+x^{2}}dx}=e^{\ln \left( 1+x^{2}\right) }=1+x^{2}[/tex] would give us anyway). So the equation becomes

[tex]\dfrac{d}{dx}\left[ \left( 1+x^{2}\right) y\right] =f(x)[/tex]

Since [tex]f[/tex] is given by two different formulas, we work on each interval separately.

On [tex]0\leq x<1[/tex], we have [tex]f(x)=x[/tex], so

[tex]\left( 1+x^{2}\right) y=\int x\,dx=\dfrac{x^{2}}{2}+C[/tex]

The initial condition [tex]y(0)=0[/tex] gives [tex]1\cdot 0=0+C\Longrightarrow C=0[/tex], so

[tex]y=\dfrac{x^{2}}{2\left( 1+x^{2}\right) }[/tex]

On [tex]x\geq 1[/tex], we have [tex]f(x)=0[/tex], so

[tex]\left( 1+x^{2}\right) y=K\Longrightarrow y=\dfrac{K}{1+x^{2}}[/tex]

The constant [tex]K[/tex] is fixed by continuity at [tex]x=1[/tex]. From the first branch [tex]y(1)=\dfrac{1}{2\cdot 2}=\dfrac{1}{4}[/tex], and from the second [tex]y(1)=\dfrac{K}{2}[/tex], so [tex]K=\dfrac{1}{2}[/tex] and

[tex]y=\dfrac{1}{2\left( 1+x^{2}\right) }[/tex]

The particular solution is therefore

[tex]y=\left\{ \begin{array}{ll} \dfrac{x^{2}}{2\left( 1+x^{2}\right) } , & 0\leq x<1 \\ \dfrac{1}{2\left( 1+x^{2}\right) } , & x\geq 1 \end{array} \right.[/tex]

Two things to watch here: the arbitrary constant belongs inside the bracket with [tex]\int f(x)dx[/tex], so it is divided by [tex]1+x^{2}[/tex] and not simply added at the end; and a particular solution contains no [tex]C[/tex] at all once the conditions have been used.