Question

For the given differential equation, find the general solution: $(1 + x^2) dy + 2xy dx = cot x dx (x \neq 0)$

Answer 1

Answer:

The solution of the given diferential equation is $y.(1+x^2)=log\:sinx+c$

Step-by-step explanation:

Concept:

The solution of

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

$(1+x^2)dy+2xy\:dx=cotx\:dx$

This can be written as

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

comparing this with

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

we get

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

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

$y.e^{\intPdx}=\int{Q.e^{\intPdx}dx}+c\\\\y.(1+x^2)=\int{\frac {cotx}{1+x^2}.(1+x^2)\:dx}+c\\\\y.(1+x^2)=\int{cotx\:dx}+c\\\\y.(1+x^2)=log\:sinx+c$