Stuck on double integral problem

jahichuanna · #1 $Old$ June 14th, 2009, 06:26 PM

${d^2y}{dx^2} = 2x-e^x$ Find y given $\frac{dy}{dx} = 1$ and y=2 when x=0

Chris L T521 · #2 $Old$ June 14th, 2009, 06:55 PM

Quote:

Originally Posted by jahichuanna $View Post$

I was studying for exams (ugh!) and came upon this problem that I need to know how to do:

$\frac{d^2y}{dx^2}=2x-e^x$ Find y given $\frac{dy}{dx} = 1$ and y=2 when x=0

I have integrated the function twice (because it's a 2nd derivative problem): $\int\int 2x-e^x = \int x^2-e^x+C = \frac{x^3}{3}-e^x+C_1x+C_2$ . But I don't know how to solve for the Cs. Can anyone offer any insight?

When you integrate the first time, you end up with $\frac{\,dy}{\,dx}=x^2-e^x+C$ . Now, you have the initial condition $y^{\prime}\!\left(0\right)=1$ . That means that $1=-1+C\implies C=2$ .

So it follows that $\frac{\,dy}{\,dx}=x^2-e^x+2$

Integrating again, we have $y=\tfrac{1}{3}x^3-e^x+2x+C$ .

Now apply the other initial condition $y\!\left(0\right)=2$ to find C.

Can you take it from here?