Question

[4] Use variation of parameters to find the general solution of each equation. (a) $y'' + 4y' + 4y = \frac{1}{x^2 e^{2x}}$ for $x > 0$ (b) $y'' - 2y' + y = \frac{e^x}{x^2 + 1}$

          [4] Use variation of parameters to find the general solution of each equation.
(a) $y'' + 4y' + 4y = \frac{1}{x^2 e^{2x}}$ for $x > 0$
(b) $y'' - 2y' + y = \frac{e^x}{x^2 + 1}$

[4] Use variation of parameters to find the general solution of each equation.
(a) y” + 4y' + 4y = (1)/(x^2 e^2x) for x > 0
(b) y” - 2y' + y = (e^x)/(x^2 + 1)

Added by Ricardo T.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Raj Bala

01/30/2022

Step 1

The characteristic equation is ar^2 + 4r + 4 = 0, which factors to (r+2)^2 = 0. This gives us a repeated root of -2, so the complementary solution is y_c = c1e^(-2x) + c2xe^(-2x). Show more…

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4a-4b, answers should be a) y=c1e^-2x + c2xe^-2x - e^-2x (lnx)b) y=c1+c2xe^x - 1/2e^x ln(x^2 +1) + xe^x tan^-1 xbut i need to see the work on how to get the answers. [4] Use variation of parameters to find the general solution of each equation ay+4y+4y= 1 forx>0 by-2y+y= x2+1