d2y/dx2-4dy/dx+3y=x+e^2x.
If Y=ae^2x, dY/dx=2ae^2x=2Y; d2Y/dx2=4ae^2x=4Y.
If z=sinx, d2z/dx2=-z and dz/dx=cosx. Since there are no trig terms in the given equation it seems unlikely that the solution involves them.
Let's assume that y=ae^2x+f(x), where f(x) is a polynomial in x and a is a constant; then dy/dx=2ae^2x+f'(x), where f' is the derivative of f, and d2y/dx2=4ae^2x+f''(x).
We can rewrite the original equation, substituting for the derivatives y' and y''.
4ae^2x+f''(x)-4(2ae^2x+f'(x))+3(ae^2x+f(x))=x+e^2x.
Let's assume that f(x)=Ax^3+Bx^2+Cx+D, where A, B, C, D are constants, then f' is 3Ax^2+2Bx+C and f'' is 6Ax+2B, so the given equation becomes:
(4ae^2x+6Ax+2B)+(-8ae^2x-12Ax^2-8Bx-4C)+(3ae^2x+3Ax^3+3Bx^2+3Cx+3D)=x+e^2x.
We can equate the coefficients on each side of the equation, so, taking e^2x coefficients we have: 4a-8a+3a=1 and -a=1 making a=-1.
There are no constants so 2B-4C+3D=0; the coefficients for x are:
6A-8B+3C=1; x^2: -12A+3B=0; x^3: 3A=0. So A=B=0 and C=1/3, D=4/9.
Therefore the solution is y=-e^2x+x/3+4/9 or 9y=3x-9e^2x+4.