Question

Solve the following equation $\dfrac{dy}{dx}-xy={{y}^{2}}{{e}^{x2}}{{/}^{2}}.\sin x$ .
A. ${{e}^{{{x}^{2}}/2}}=y\left( c+\cos x \right)$.
B. ${{e}^{-{{x}^{2}}/2}}=y\left( c+\cos x \right)$.
C. ${{e}^{{{x}^{2}}}}^{+2}=y\left( c-\cos x \right)$.
D. ${{e}^{-{{x}^{2}}/2}}=y\left( c-\cos x \right)$.

Answer 1

Hint: At first take y as $\dfrac{1}{4}$ and then transform the equation. Also substitute $\dfrac{dy}{dx}$ as $\dfrac{-1}{{{4}^{2}}}\dfrac{dy}{dx}$. Then multiply by ${{4}^{2}}$ throughout the equation and thus further solve the differential equation.

Complete step-by-step answer:
In the question a differential equation is given $\dfrac{dy}{dx}-xy={{y}^{2}}{{e}^{{{x}^{2}}/2}}\sin x$ and we have to find the original equation.
The given differential equation is, $\dfrac{dy}{dx}-xy={{y}^{2}}{{e}^{{{x}^{2}}/2}}\sin x$ .
It is in the form of Bernoulli equation which is represented as,
$\dfrac{dy}{dx}+P\left( x \right)y=Q\left( x \right){{y}^{n}}.$
Where n is the real number but not be 0 or 1 and $P\left( x \right)$ and $Q\left( x \right)$ are referred to as functions of x.
Now to solve it we have substituted ${{y}^{1-n}}$ as 4.
Here the value of n is 2 so, 4 is equal to $\dfrac{1}{y}$ .
So instead of substituting $\dfrac{1}{y}$ as we will substitute y as $\dfrac{1}{4}$ .
So we can write it as,
$\dfrac{dy}{dx}-\dfrac{x}{4}=\dfrac{{{e}^{{{x}^{2}}/2}}}{{{4}^{2}}}$ .
As $y=\dfrac{1}{4}$ so, $\dfrac{dy}{dx}=\dfrac{-1}{{{4}^{2}}}\dfrac{dy}{dx}$ .
So, we will substitute $\dfrac{dy}{dx}$ as $\dfrac{-1}{{{4}^{2}}}\dfrac{dy}{dx}$. In the equation and get, $\dfrac{-1}{{{4}^{2}}}\dfrac{dy}{dx}-\dfrac{-x}{4}=\dfrac{{{e}^{-{{x}^{2}}/2}}\sin x}{{{4}^{2}}}$ .
Now by multiplying by $-{{4}^{2}}$ we get,
$\dfrac{dy}{dx}+4x=-{{e}^{-{{x}^{2}}/2}}\sin x$ .
So, the new equation in terms of 4 , x is,
$\dfrac{dy}{dx}+4x=-{{e}^{-{{x}^{2}}/2}}\sin x$ .
Now to solve the equation we have to multiply by integration factor which is ${{e}^{\int P\left( x \right)dx}}$
Here $P\left( x \right)$ is x and $Q\left( x \right)$ is ${{e}^{-{{x}^{2}}/2}}\sin x$ .
So the integration factor is ${{e}^{\int xdx}}$ or ${{e}^{{{x}^{2}}/2}}$ .
Now we will multiply by ${{e}^{{{x}^{2}}/2}}$ through the equation so we get,
${{e}^{{{x}^{2}}/2}}\dfrac{dy}{dx}+4x{{e}^{{{x}^{2}}/2}}=-{{e}^{-{{x}^{2}}/2}}\sin x-{{e}^{{{x}^{2}}/2}}$ or, ${{e}^{{{x}^{2}}/2}}\dfrac{dy}{dx}+4x{{e}^{{{x}^{2}}/2}}=-\sin x$.
So, we can further rewrite equation as, ${{e}^{{{x}^{2}}/2}}\dfrac{d\left( 4 \right)}{dx}+4.\dfrac{d}{dx}\left( {{e}^{{{x}^{2}}/2}} \right)=-\sin x$ or, $\dfrac{d}{dx}\left( 4.{{e}^{{{x}^{2}}/2}} \right)=-\sin x$
So, we can write it as, $d\left( 4.{{e}^{{{x}^{2}}/2}} \right)=-\sin xdx$.
Now on integrating on both the sides we get, $\int d\left( 4.{{e}^{{{x}^{2}}/2}} \right)=\int -\sin xdx$ .
So, on integrating we get,
$4.{{e}^{{{x}^{2}}/2}}=c+\cos x$ .
Here c is constant of integration.
Now as we took $\dfrac{1}{y}$ as 4 in the beginning so we will write as,
$\dfrac{1}{y}{{e}^{{{x}^{2}}/2}}=c+\cos x$ or, ${{e}^{{{x}^{2}}/2}}=y\left( c+\cos x \right)$ .
Hence the answer is $'A'$.

Note: In the Bernoulli equation used, the value of n cannot be 0 or 1 because when $n=0$ the equation will be solved by first or linear differential equation and if $n=1$ then it will be solved by using separation of variables.