Differential equation:

Question

Solve y"-2y'+5y = e^2x

Answer 1

We have

(D² – 2D + 5)y = e^2x

A.E. is m² - 2m + 5 = 0

m = 1 \(\pm\) 2i

C.F. = e^x (C₁ cos 2x + C₂ sin 2x)

Here \(\phi\)(x) = e^2x and 2 is not a root of the A.E.

We assume for P.I. in the form y_p = ae^2x ...(1)

We have to find 'a' such that

y''_p - 2y'_p + 5y_p = e^2x ......(2)

From Eqn. (1),

y'_p = 2a e^2x

y''_p = 4a e^2x

Eqn. (2),

⇒ 4a e^2x - 4a e^2x + 5a e^2x = e^2x

⇒ 5a e^2x = e^2x

Equating the coefficients

5a = 1

a = \(\frac 15\)

Eqn. (1),

y_p = \(\frac 15\) e^2x

\(\therefore\) y = C.F. + P.I.

y = e^x (C₁ cos 2x + C₂ sin 2x) + \(\frac 15\) e^2x