y''=d²y/dx², y'=dy/dx for ease of writing

First off split the full equation into th homogeneous equation and the particular function.

The homogenous equatiion is when our collection of y, y', y'' are on one side and equal to 0 so; y''+2y'-3y=0.

When first looking at a question like this our initial intuition should be to think of a function, y, whose derivatives are similar but not equal to itself. y=Keⁿ^x is such a function as its derivatives are y'=Kneⁿ^x, y''=Kn²e^nx

When we substitue this into out homogeneous equation we get:

Ke^nx(n²+2n-3)=0

and as e^nx can never equal zero we have : n²+2n-3=0

This is called a characteristic function and in the exam you just need to be anle to find this.

from this we get the roots n=1,n=-3.

Substituting this back into out guessed function for y we get that y₁=Ae^x and y₂=Be^-3x where A and B are constants.

We can then combine these two independant solutions to our homogeneous equation to form a complimentary function:

y_cf=y₁+y₂=Ae^x+Be^-3x

We can then deal with our particular function. Again we guess what might work in this case probably a polynomial.

So: y=ax+b

y'=a

y''=0

Substituting into our original equation we get:

0+2a-3ax-3b=x

comparing coefficients gives us:

a=-1/3

b=-2/9

so our particular function y_p=-x/3 -2/9

Our general solution consists of our solution to the homogeneuos case (where it equals zero) and the particular solution. Because the homogeneous solution is the solution for the set of differential equations which start with y''+2y'-3y by adding the particular solution for x we get our general solution.

So y=y_cf+y_p= Ae^x+Be^-3x -x/3 -2/9

Find the determinant of a 3x3 matrix.

Related Further Mathematics A Level answers

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