For the differential equation y\" + 4y\' + 4y = 0 find the roots of the characte

Question

For the differential equation y" + 4y' + 4y = 0 find the roots of the characteristic equation. Enter the roots as a comma separated list, e.g. 2, -7 or 1 -2i, 1 + 2i. In case of a double root list the root twice. Roots: r = The general solution (here r0, r1, r2, alpha, beta are real numbers) can be written in one of the the forms c1er1x + c2er2x c1er0x + c2xer0x c1e alpha x cos(beta x) + c2e alpha x sin(beta x) Using parameters c1 and c2 for c1 and c2 write the general solution, e.g., you could write something like c1 e^ (3x) +c2 e^ (-x) y(x) =

Explanation / Answer

Here the main differential equation will become

D^2 + 4D + 4 = 0

(D+2)^2 = 0

Therefore, (D + 2)(D + 2) = 0

Therefore, roots are -2 and -2

Final solution of the differential equation will be

y = c1*e^(-2x) + c2* x * e^(-2x)

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