Question

The solution of the differential equation $x\dfrac{dy}{dx}+2y={{x}^{2}}\left( x\ne 0 \right)$ with $y\left( 1 \right)=1,$ is:
$\left( a \right)y=\dfrac{{{x}^{3}}}{5}+\dfrac{1}{5{{x}^{2}}}$
$\left( b \right)y=\dfrac{4}{5}{{x}^{3}}+\dfrac{1}{5{{x}^{2}}}$
$\left( c \right)y=\dfrac{3}{4}{{x}^{2}}+\dfrac{1}{4{{x}^{2}}}$
$\left( d \right)y=\dfrac{{{x}^{2}}}{4}+\dfrac{3}{4{{x}^{2}}}$

Answer 1

We are given the differential equation as $x\dfrac{dy}{dx}+2y={{x}^{2}}.$ We will convert it into a linear form $\dfrac{dy}{dx}+Py=Q\left( x \right)$ and then find the integrating factor using $I.F={{e}^{\int{Pdx}}}.$ Once we have our integrating factor, we will find the solution using $y\times IF=\int{\left( Q\times IF \right)dx+C}.$ We will put the value of Q and IF and solve for y which is our required solution.

Complete step-by-step answer:
We are given the differential equation as $x\dfrac{dy}{dx}+2y={{x}^{2}}.$ First, we will convert our differential equation into the linear form. We have
$x\dfrac{dy}{dx}+2y={{x}^{2}}$
Dividing both the sides by x, we get,
$\Rightarrow \dfrac{dy}{dx}+\dfrac{2y}{x}=x$
So, we get our linear form. So comparing with $\dfrac{dy}{dx}+Py=Q,$ we get, $P=\dfrac{2}{x}$ and Q = x.
Now, to solve further we will find the integrating factor. The integrating factor is given as
$I.F={{e}^{\int{Pdx}}}$
As, $P=\dfrac{2}{x},$ so we get,
$\Rightarrow I.F={{e}^{\int{\dfrac{2}{x}dx}}}$
As, $\int{\dfrac{1}{x}dx}=\log x,$ so we get,
$\Rightarrow IF={{e}^{2\log x}}$
As, $n\log x=\log {{x}^{n}},$ so we get,
$\Rightarrow IF={{e}^{\log {{x}^{2}}}}$
Now, we know that, ${{e}^{\log x}}=x,{{e}^{\log {{x}^{2}}}}={{x}^{2}},$ so we get,
$\Rightarrow IF={{x}^{2}}$
Now, we know the solution is given as,
$y\times IF=\int{\left( Q\times IF \right)dx+C}$
As, $IF={{x}^{2}},Q=x,$ so we get,
$\Rightarrow y{{x}^{2}}=\int{\left( x\times {{x}^{2}} \right)dx}$
$\Rightarrow y{{x}^{2}}=\int{{{x}^{3}}dx}$
As $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n},$ so,
$\Rightarrow y{{x}^{2}}=\dfrac{{{x}^{4}}}{4}+C$
Dividing both the sides by ${{x}^{2}},$ we get,
$\Rightarrow y=\dfrac{{{x}^{2}}}{4}+\dfrac{C}{{{x}^{2}}}$
Now, we have y(1) as 1, i.e y (1) = 1. So, putting x = 1 and y = 1, we get,
$\Rightarrow 1=\dfrac{{{1}^{2}}}{4}+\dfrac{C}{{{1}^{2}}}$
$\Rightarrow 1=\dfrac{1}{4}+\dfrac{C}{1}$
$\Rightarrow 1=\dfrac{1}{4}+C$
Solving for C, we get,
$\Rightarrow C=1-\dfrac{1}{4}=\dfrac{3}{4}$
Now, we get, $C=\dfrac{3}{4}.$
So, putting it in the value of y, we get,
$\Rightarrow y=\dfrac{{{x}^{2}}}{4}+\dfrac{3}{4{{x}^{2}}}$

So, the correct answer is “Option d”.

Note: Remember that while finding the integrating factor, we apply the differentiation. First, we integrate P with respect to x and then solve it over the exponential. As we do that we have P as $\dfrac{2}{x}.$ So,
$\int{\dfrac{2}{x}dx}=2\int{\dfrac{1}{x}dx}$
$=2\log x$
Then we apply ${{e}^{\int{\dfrac{2}{x}dx}}}={{e}^{2\log x}}.$ Also, $n\log x=\log {{x}^{n}}.$ So, we get, ${{e}^{\log {{x}^{2}}}}$ and then we simplify and get IF as ${{x}^{2}}.$ So, we do it by part by part so that no errors occur.