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Solve the differential equation y^(4) + 4y'' + 4y = 0 with initial conditions y(0) = 1, y'(0) = 0, y''(0) = -2, y'''(0) = 0.

April 10, 2026Mathematics

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Step-by-step solution

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1. Solve the following differential equation:
$y^{(4)} + 4y'' + 4y = 0$
Find the general solution and then determine the particular solution satisfying $y(0) = 1$ , $y'(0) = 0$ , $y''(0) = -2$ , $y'''(0) = 0$ .

Find the characteristic equation.
The given differential equation is $y^{(4)} + 4y'' + 4y = 0$ .
The characteristic equation is obtained by replacing derivatives with powers of $r$ :
$r^4 + 4r^2 + 4 = 0$
Solve the characteristic equation.
Let $u = r^

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Solve the differential equation y^(4) + 4y'' + 4y = 0 with initial conditions y(0) = 1, y'(0) = 0, y''(0) = -2, y'''(0) = 0.

Step-by-step solution

Find the characteristic equation.

Solve the characteristic equation.

Still stuck on this one?

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