Find the general solution for differential equations. (1 + x) dy\dx - y =e3x (1 + x)2

Question

Find the general solution for differential equations.

\((1+x)\frac{dy}{dx} - y= e^{3x}\,(1+x)^2\)

(1 + x) dy\dx - y =e^3x (1 + x)²

Answer 1

Given Differential Equation :

\((1+x)\frac{dy}{dx} - y= e^{3x}\,(1+x)^2\)

Formula :

i) \(\int \frac{1}{px+q}\, dx = \frac{1}{p}\) log (px+ q)

ii) alog b = log b^a 

iii) a^{log_a b} = b

iv) \(\int e^{kx} \, dx = \frac{1}{k}\,e^{kx}\) 

v) General solution :

For the differential equation in the form of

\(\frac{dy}{dx} + Py = Q\)

General solution is given by,

y. (I. F.) = \(\int\) Q. (I. F.) dx + c

Where, integrating factor,

I. F. = \(e^{\int p\,dx}\) 

Given differential equation is

\((1+x)\frac{dy}{dx} - y= e^{3x}\,(1+x)^2\)

Dividing above equation by (1+x),

\(\frac{dy}{dx} - \frac{1}{(1+x)}\, y = e^{3x}\,(1+x)\) …eq(1)

Equation (1) is of the form

\(\frac{dy}{dx} + Py = Q\)

Where, \(P = \frac{-1}{(1+x)}\,and\,Q = e^{3x}\,(1+x)\)

Therefore, integrating factor is

General solution is

Multiplying above equation by (1+x),

∴ y = \(\frac{1}{3} (1 + x)\, e^{3x} + c\,(1+x)\)

Therefore general equation is

 y = \(\frac{1}{3} (1 + x)\, e^{3x} + c\,(1+x)\)