Hi, can someone please help me to solve this?
y''+9y+20=0,
y(0)=1, y'(0)=1
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 $\displaystyle y=-{\textstyle\frac{29}2}x^2 + x + 1$ then $\displaystyle y' = -29x + 1$, and $\displaystyle y''=-29$. If you plug these formulas for y" and y into the expression $\displaystyle y'' + 9y + 20$, 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 $\displaystyle y'' + 9y = -20$, 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 $\displaystyle y'' + 9y = 0$ is $\displaystyle y = A\cos 3x + B\sin 3x$. 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 $\displaystyle \textstyle y = -\frac{20}9$.
Add the solution to the homogeneous equation to the particular solution, and you get $\displaystyle y = A\cos 3x + B\sin 3x - {\textstyle\frac{20}9}$. 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.)