Solving y'= 3y + 15: Initial Value Problem

[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!

 

[Sirus]

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

 

[vsage]

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!

This problem is easily solved by separation 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?

 

[Sirus]

Ofcourse! It's been too long...

 

[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.