Find the particular solution, y=f(x), to the differential equation (1+x^2 ) dy/dx=x^3/y where f(0)=-3.

Question

1. Find the particular solution,  to the differential equation  where

 

Answer 1

(1+x^2)dy/dx=x^3/y; ydy=(x^3/(1+x^2))dx=(x-x/(1+x^2))dx.

∫ydy=∫(x-(1/2).2x/(1+x^2))dx=x^2/2-(1/2)ln(1+x^2).

So y^2=x^2-ln(1+x^2)+C; y=-3 when x=0, so 9=C and y^2=x^2-ln(1+x^2)+9.

f(x)=-√(x^2-ln(1+x^2)+9), so f(0)=-3.

Answer 2

Find the particular solution, y=f(x), to the differential equation (1+x^2 ) dy/dx=x^3/y where f(0)=-3.

(1+x^2 ) dy/dx = x^3/y

Cross-multiplying,

y dy/dx = x^3/(1 + x^2)

y dy = x^3/(1 + x^2) dx

int y dy = int x^3/(1 + x^2) dx = int (x + x^3)/(1 + x^2) - x/(1 + x^2) dx

int y dy = int x dx – int x/(1 + x^2) dx

y^2/2 = x^2/2 – (1/2)ln(1 + x^2) + C

y^2 = x^2 – ln(1 + x^2) + D

y = √(x^2 – ln(1 + x^2) + D)

Using the initial condition, f(0) = -3,

-3 = √(0 - ln(1) + D)

D = 9

The particular solution is.

f(x) = √( x^2 – ln(1 + x^2) + 9)