Answer to Question #178362 in Differential Equations for Hassan Jahangir

Question

Solve the following non-homogeneous linear ODE of first order

dy/dx + 3y/3x = 6x^2

 

Answer 1

Given,

$\dfrac{dy}{dx}+\dfrac{3y}{3x}=6x^2$

This is linear differential equation so

Here $P=\dfrac{1}{x}, Q=6x^2$

Integrating factor $I.F.=e^{\int Pdx}=e^{\dfrac{1}{x}dx}=e^{logx}=x$

Therefore the solution is-

$y\times I.F.=\int Q\times I.F. dx$

$y\times x=\int 6x^3dx$

$y\times x=\dfrac{6x^4}{4}+C$

$yx=\dfrac{3}{2}x^4+C$

Hence, $y=\dfrac{3}{2}x^3+\dfrac{C}{x}$