Question

For the given differential equation, find the particular solution satisfying the given condition: [tex](1+x^2)\frac{dy}{dx} +2xy=\frac {1}{1+x^2}; y = 0\ when x = 1

Answer 1

Answer:

Step-by-step explanation:

Concept:

The solution of

$\frac{dy}{dx}+Py=Q\: is\\\\y.e^{\int{P}dx}=\int{Q.e^{\intPdx}dx}+c$

$(1+x^2)\frac{dy}{dx} +2xy=\frac {1}{1+x^2}$

divide both sides by $1+x^2$

$\frac{dy}{dx} +\frac{2xy}{1+x^2}=\frac {1}{(1+x^2)^2}$

comparing this with

$\frac{dy}{dx}+Py=Q$

we get

$P=\frac{2x}{1+x^2}\\\\Q= \frac {1}{(1+x^2)^2}$

Integrating factor

$=e^{\int{P}dx}\\\\=e^{\int{\frac{2x}{1+x^2}}\:dx}\\\\=e^{log(1+x^2)}\\\\=1+x^2$

The solution is

$y.e^{\int{P}dx}=\int{Q.e^{\intPdx}dx}+c\\\\y.(1+x^2)=\int{\frac {1}{(1+x^2)^2}.(1+x^2)dx}+c\\\\y.(1+x^2)=\int{\frac {1}{1+x^2}dx}+c\\\\y.(1+x^2)=tan^{-1}x+c$

y=o when x=1,

$0=tan^{-1}1+c\\\\0=\frac{\pi}{4}+c\\\\c=-\frac{\pi}{4}$

The required solution is

$y.(1+x^2)=tan^{-1}x-\frac{\pi}{4}$