Question

Will give medal and. Fan.
Find the initial value problem of
d^2y/dx ^2 + x dy/dx + y=0 given that y(0)=1 and dy/dx (0)=0

Answer 1

Notice that
\[\frac{d^2y}{dx^2}=\frac{d}{dx}\left[\frac{dy}{dx}\right]\quad\text{and}\quad x\frac{dy}{dx}+y=\frac{d}{dx}\left[xy\right]\]
So, the original ODE is equivalent to
\[\frac{d}{dx}\left[\frac{dy}{dx}+xy\right]=0\]
Integrating both sides yields
\[\frac{dy}{dx}+xy=C\]
and you have an equation linear in \(y\). Find your integrating factor.
\[\ln\mu(x)=\int x\,dx~~\implies~~\mu(x)=e^{x^2/2}\]
Now solve,
\[e^{x^2/2}\frac{dy}{dx}+xe^{x^2/2}y=Ce^{x^2/2}\]

Answer 2

please can you help me finish it sir.
it is an exam question . i dont want to do any error

Answer 3

help me sir