a) Find the general solution to the differential equation: f(x)=y''-12y'-13y=8. b) Given that when x=0, y=0 and y'=1, find the particular solution to f(x).

First consider y''-12y'-13y=0.

Try: y=e^mx. This gives y'=me^mx and y''=m²e^mx.

Substituting into the differntial equation we get: m²e^mx-12me^mx-13e^mx=0

Since e^mx>0 divide throughout by e^mx to get: m²-12m-13=0

This can be factorised into (m-13)(m+1)=0 so the roots of the quadratic equation are: m1=13 m2=-1.

Therefore the complementary solution is Ae^13x+Be^-x.

he particular integral is in the form of λ.

So let y=λ, which gives y'=0 and y''=0 since λ is a constant.

Substituting into the differential equation we get 0+0-13λ=8 so λ=-8/13.

We know that general solution=complementary function + particular integral.

So the general solution is: y=Ae^13x+Be^-x-8/13.

b) We will now use the initial conditions given to find the particular solution.

We are given two initial conditions since this is a second order differential equation.

Using the fact that when x=0, y=0, we get: 0=A+B-8/13 therefore A+B=8/13.

Rearranging for B we get: B=8/13-A.

To use the fact that x=0, y'=0 we need to differentiate y.

y'=13Ae^13x-Be^-x. So 1=13A-B.

Rearranging for B we get: B=13A-1=8/13-A (from the first expression for B).

So 13A+A=8/13+1, so 14A=21/13, so A=3/26 and B=13 *3/26 -1= 3/2 -1=1/2.

Substituting these into the general solution, we get that the particular solution to f(x) is y=3/26e^13x+1/2e^-x-8/13.

If you have time in the end, substitute this back into f(x) to double check your answer.