# Finding a solution?

1. Sep 25, 2004

### parwana

Finding a solution??

Find all solutions to

y'= 3y + 15

and solve the initial value problem-
y'=3y+15
y(0)= -1

so would it be y= 3/2x^2 + 15y + C?????

if not please share what would it be!!!!

2. Sep 25, 2004

### Sirus

This should be in Calculus and Analysis. (sorry, I'm not sure about your quesiton.)

Last edited: Sep 25, 2004
3. Sep 25, 2004

### vsage

This problem is easily solved by seperation of variables (which I see you did). However, I'm pretty sure you meant 3/2y^2. if y(0) = -1, what does that tell you about the value of C?

4. Sep 25, 2004

### Sirus

Ofcourse! It's been too long...

5. Sep 25, 2004

### BLaH!

This is a first order linear differential equation. It will be instructive to write this equation as $$\frac{dy}{dx} = 3y + 15$$. Because the right side contains $$y(x)$$ you can't directly integrate the equation as it looks like you tried to do. Instead you can write it so all the $$y$$ stuff is on the left and all the $$x$$ is on the right: $$\frac{dy}{3y + 15} = dx$$. NOW we can integrate both sides to get $$\int_{-1}^{y}\frac{dy}{3y + 15} = \int_{0}^{x}dx$$. Notice that I've already included the initial condition that $$y = -1$$ when $$x=0$$ in my limits of integration. It is now a simple matter of integrating the left side and then solving for $$y(x)$$. The answer I got was $$y(x) = 4e^{3x}-5$$.

Last edited: Sep 26, 2004