Thread: Easy Problem made frustrating

d^2y/dx^2 = 0.5dy/dx where y=0 and dy/dx=3 when x=0
We have to find what y is when x=5
My solution keeps turning out to be y=6e^5/2 - 6
However the back of my textbook says y=6e^5/2 - 6e^2
Here is how i did it
let v = dy/dx
therefore d^2y/dx^2 = vdv/dy
vdv/dy = 0.5v
dv=0.5dy ... intergrate both sides
v=0.5y + C
3=0.5(0) + C
C=3
v=0.5y + 3
dy/dx=0.5y + 3
2dy/y+6 = dx
2loge(y+6) = x + B
B = 2loge(6)
2loge(y+6/6)=x
y+6/6=e^x/2
y=6e^5/2 - 6
Where have I gone wrong and what is the correct way to do it if i am wrong???

Ok, so 2nd derivative is .5 times first derivative, means if that was your differential equation, then y=C1*e^(.5*x) is one fundamental solution, and y=C2 is the other. At x=0, there are the boundary conditions y=0 and y'=3. You stated it really badly by the way. The first few times I read it, I thought you meant d^2y/dx^2 = 0.5dy/dx was true only at y=0, and dy/dx=3 is true when x=0, which would be 2 boundary conditions and no differential equation. So obviously C1=-C2 so that y(0)=0. Then y'=.5*C1=3 at x=0, so C1=6. y=6*(e^(.5*x)-1). If x=5, then y=6*(e^(2.5)-1). Apparently you didn't go wrong, since you got the same answer I did, although I never involved the natural log.

sorry about mix up, thks for clearing this up, i don't understand your method, (i'm pretty useless), but i'm relieved you got the same answer. now i can sleep in peace lol.