Answer to Question #262005 in Differential Equations for anje

Question

Obtain the particular solution

1. dy/dx+2y=y^3e^4x, y(0)=1. ans. y^2 (1-2x)=e^-4x

2.dy/dx - y/x=y^5/x^3 ,y(1)=-1. ans. y^4(3-2x^2)=x^4

Answer 1

1.

$z=y^{1-3}=y^{-2}$

$\dfrac{dz}{dx}=-2y^{-3}\dfrac{dy}{dx}$

$-2y^{-3}\dfrac{dy}{dx}-2y^{-3}(2y)=-2y^{-3}y^3e^{4x}$

$\dfrac{dz}{dx}-4z=-2e^{4x}$

Integration factor

$\mu(x)=e^{-4x}$

$e^{-4x}\dfrac{dz}{dx}-4e^{-4x}z=-2e^{-4x}e^{4x}$

$d(e^{-4x}z)=-2dx$

Integrate

$\int d(e^{-4x}z)=-\int2dx$

$e^{-4x}z=-2x+C$

$y^{-2}=-2xe^{4x}+Ce^{4x}$

$y^2=\dfrac{1}{-2xe^{4x}+Ce^{4x}}$

$y(0)=1$

$1^2=\dfrac{1}{-2(0)e^{4(0)}+Ce^{4(0)}}=>C=1$

$y^2(1-2x)e^{4x}=1$

$y^2(1-2x)=e^{-4x}$

2.

$z=y^{1-5}=y^{-4}$

$\dfrac{dz}{dx}=-4y^{-5}\dfrac{dy}{dx}$

$-4y^{-5}\dfrac{dy}{dx}+4y^{-5}(\dfrac{y}{x})=-4y^{-5}\dfrac{y^5}{x^3}$

$\dfrac{dz}{dx}+4\dfrac{z}{x}=-\dfrac{4}{x^3}$

Integration factor

$\mu(x)=e^{\int(4/x)dx}=x^{4}$

$x^{4}\dfrac{dz}{dx}-4x^{4}\dfrac{z}{x}=-x^{4}\dfrac{4}{x^3}$

$d(x^{4}z)=-4xdx$

Integrate

$\int d(x^{4}z)=-\int4xdx$

$x^{4}z=-2x^2+C$

$z=-2x^{-2}+Cx^{-4}$

$y^{-4}=-2x^{-2}+Cx^{-4}$

$y(1)=-1$

$(-1)^{-4}=-2(1)^{-2}+C(1)^{-4}=>C=3$

$y^{-4}=-2x^{-2}+3x^{-4}$

$y^4(3-2x^2)=x^4$