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

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

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

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

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

$y(1)=-1$

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

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

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