Solve y''-2y'+5y=(x^2)+1

Question

Solve. y'' - 2y' + 5y = (x^2) + 1

Answer 1

First, solve the homogenous equation y'' -2y' + 5y = 0, which has constant coefficients, so finding its solutions reduces to finding the roots of the quadratic polynomial r²-2r+5=0. Using the quadratic formula we get roots 1±2i, so the homogeneous solution is: y_h= C₁e^x cos(2x) + C₂e^x sin(2x).

Now, to find a particular solution, we can use the method of undetermined coefficients. That is, we seek one of the form y_p= Ax²+ Bx + C, because the right-hand-side of the equation is quadratic (and also the homogeneous solution does not have this form). The derivatives of y_p are

y_p' = 2Ax + B, and y_p'' = 2A. Plug them into the original equation and get:

2A -2(2Ax + B) +5(Ax²+ Bx + C) = x²+1, or equivalently

5A x² + (-4A + 5B)x + 2A - 2B + 5C = x²+1

Equating the coefficients we get equations

5A = 1 → A = 1/5

-4A + 5B = 0 → B = 4A/5 = 4/25

2A - 2B + 5C = 1 → C = (1-2A + 2B)/5 = (1-2/5 + 8/25)/5 = 23/125

so y_p = x²/5 + 4x/25 + 23/125

Altogether, the general solution is y = C₁e^x cos(2x) + C₂e^x sin(2x) + x²/5 + 4x/25 + 23/125