Question

For the given differential equation, find the general solution: $\frac{dy}{dx} + 2y= sin\ x$

Answer 1

Answer:

The required solution is

$y.e^{2x}=\frac{e^{2x}}{5}[2sinx-cosx]+c$

Step-by-step explanation:

Concept:

The solution of

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

$\frac{dy}{dx}+2y=sin\:x$

Comparing this equation with

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

we get

P=2, Q=sinx

Integrating factor

$=e^{\int{P}\:dx}\\\\=e^{\int2\:dx}\\\\=e^{2\int\:dx}\\\\=e^{2x}$

Solution:

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

$y.e^{2x}=\int{e^{2x}sinx\:dx}+c$

$y.e^{2x}=\frac{e^{2x}}{2^2+1^2}[2sinx-cosx]+c$

$y.e^{2x}=\frac{e^{2x}}{5}[2sinx-cosx]+c$