Solve 1x3dydx3x2yfx explicitly with initial condition y01 an

Solve (1+x^3)(dy/dx)+3x^2y=f(x) explicitly with initial condition y(0)=1 and f(x)={e^x, 0<=x<=2; -x, x>=2

Solution

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

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

IF = e^(integral of (3x^2 / 1 + x^3) )

IF = e^ln|1 + x^3|

IF = 1 + x^3

Multiply all over by 1 + x^3 :

(1 + x^3) * [dy/dx + (3x^2 / 1 + x^3) * y = f(x) / (1 + x^3)]

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

d/dx(y(1+x^3)) = f(x)

Integrating :

y(1+x^3) = integral of f(x)

f(x)={e^x, 0<=x<=2;
-x, x>=2 }

y(0) = 1

So, f(x) = e^x

y(1 + x^3) = integral of e^x

y(1 + x^3) = e^x + C

y(0) = 1 :

1(1 + 0^3) = e^0 + C

1 = 1 + C

C = 0

So, y(1 + x^3) = e^x

y = e^x / (1 + x^3) ---> ANSWER