Answer to Question #223123 in Differential Equations for Caren

Question

Find the particular solution of each of the differential equation expressing y explicitly in terms of x.

a) y²dy/dx = 2x²+1 y=1 when x=1

b) xdy/dx = y+2 y=7 when x=3

Answer 1

a) Integrate

$$\int y^2dy=\int (2x^2+1)dx$$

$$\dfrac{1}{3}y^3 =\dfrac{2}{3}x^3+x+\dfrac{1}{3}C$$

$$y^3 =2x^3+3x+C$$

$y(1)=1$

$$(1)^3 =2(1)^3+3(1)+C$$

$$C=-4$$

The particular solution of each of the given differential equation is

$$y=\sqrt[3]{2x^3+3x-4}$$

b)

$$xdy/dx = y+2$$

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

Integrate

$$\int \dfrac{dy}{y+2}=\int\dfrac{dx}{x}$$

$$\ln(|y+2|)=\ln(|x|)+\ln C$$

$$y+2=Cx$$

$y(3)=7$

$$7+2 =C(3)=>C=3$$

$$C=-4$$

The particular solution of each of the given differential equation is

$$y=3x-2$$