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

$$y = -1$$What 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! :)