# Math Help - Solving a begin value problem

1. ## Solving a begin value problem

y''+9y+20=0,
y(0)=1, y'(0)=1

2. This is a very standard initial value problem. Have you been taught how to solve these? If so, where are you having difficulties with this one? Show us your working so we know where you need help.

3. Thank you for your quick reply.

The textbook I have explains a different type of 'initial value problem', so I'm not really sure that the way I tried to solve the problem is allowed. This is what I have so far:

4. Sorry to have to tell you, but this is way off the mark. You can always check the answer to a differntial equation problem by seeing whether it satisfies the given equation, and this one certainly does not. If $y=-{\textstyle\frac{29}2}x^2 + x + 1$ then $y' = -29x + 1$, and $y''=-29$. If you plug these formulas for y" and y into the expression $y'' + 9y + 20$, you certainly don't get 0.

The way to solve an equation of this sort is to forget about the initial conditions y(0)=1 and y'(0)=1 entirely, until the very end of the solution.

Write the differential equation as $y'' + 9y = -20$, and look first at the so-called homogeneous equation that you get when you replace the -20 on the right-hand side by 0. The general solution to the equation $y'' + 9y = 0$ is $y = A\cos 3x + B\sin 3x$. You then have to look for a particular solution to the equation with the -20 on the right-hand side. There is an obvious solution in this case, namely to take y to be the constant function $\textstyle y = -\frac{20}9$.

Add the solution to the homogeneous equation to the particular solution, and you get $y = A\cos 3x + B\sin 3x - {\textstyle\frac{20}9}$. This is the general solution to your equation. To finish the job, you have to choose the constants A and B so that the initial conditions y(0)=1 and y'(0)=1 are satisfied. (And if you want to finish the job properly, you should then differentiate your answer twice, and check that it really does satisfy the differential equation.)