Question

3. Solve the following: (a) y" - 2y' + 2y - e$^x$ sec(x) = 0, using the variation of parameter method. (b) x$^3$y" + x$^2$y' - 4xy - 1 - x cos(ln x) = 0. (c) y" - 4y' + 4y = xe$^{-2x}$ + e$^{2x}$ by operator method.

          3. Solve the following:
(a) y" - 2y' + 2y - e$^x$ sec(x) = 0, using the variation of parameter method.
(b) x$^3$y" + x$^2$y' - 4xy - 1 - x cos(ln x) = 0.
(c) y" - 4y' + 4y = xe$^{-2x}$ + e$^{2x}$ by operator method.

Added by Michael L.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Adi S

07/15/2023

Step 1

The characteristic equation is r^2 - 2r + 2 = 0. Using the quadratic formula, we find that the roots are r = 1 ± i. Therefore, the complementary solution is y_c = c1e^x cos(x) + c2e^x sin(x), where c1 and c2 are arbitrary constants. Show more…

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3. Solve the following: (a) y" -- 2y' + 2y -- e" sec(x) = 0, using the variation of parameter method. (b) x3y"+x2y'-4xy-1-x cos(ln x)=0. (c) y" - 4y'+ 4y = xe-2x + e2 by operator method.