Differential Equations: Solving with Two Methods

[stefan10]

Homework Statement

Solve each of these differential equations by two different methods.

\frac{dy}{dx} = 4(y+1)x^3

Homework Equations

Integrating factor

\rho = \exp (\int(p(x) dx)

Linear Equation

\frac{dy}{dx} + p(x) y(x) = Q(x)

The Attempt at a Solution

So I first solved it using separation. I get y=A \exp(x^4) -1

I then try to solve it using Integrating factors.

\frac{dy}{dx} + - 4x^3 y=4x^3
\rho = \exp (\int -4x^3 dx) = exp(-x^4)

Multiplying through by the integrating factor I get.

\exp(-x^4) y= \int 4x^3 exp(-x^4) dx

but when I solve it, I get

y = -1What can I be doing wrong? I checked my steps multiple times.

 

[Dick]

Homework Statement

Solve each of these differential equations by two different methods.

\frac{dy}{dx} = 4(y+1)x^3

Homework Equations

Integrating factor

\rho = \exp (\int(p(x) dx)

Linear Equation

\frac{dy}{dx} + p(x) y(x) = Q(x)

The Attempt at a Solution

So I first solved it using separation. I get y=A \exp(x^4) -1

I then try to solve it using Integrating factors.

\frac{dy}{dx} + - 4x^3 y=4x^3
\rho = \exp (\int -4x^3 dx) = exp(-x^4)

Multiplying through by the integrating factor I get.

\exp(-x^4) y= \int 4x^3 exp(-x^4) dx

but when I solve it, I get

y = -1


What can I be doing wrong? I checked my steps multiple times.

You are forgetting the constant of integration on $\int 4x^3 exp(-x^4) dx$.

 

[stefan10]

You are forgetting the constant of integration on $\int 4x^3 exp(-x^4) dx$.

Oh yeah! I knew it was something so simple. Thank you very much! :)