What is the solution for y^{''}-4y^'+3y=10cos(x) ?

The solution for y^{''}-4y^'+3y=10cos(x) is y=c_{1}e^{3x}+c_{2}e^x-2sin(x)+cos(x)

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y^{''}-4y^'+3y=10cos(x)

y^{^{''}} - 4 y^{^{'}} + 3 y = 10 cos (x)

y = c_{1} e^{3 x} + c_{2} e^{x} - 2 sin (x) + cos (x)

Solution steps

y^{''} (x) - 4 y^{'} (x) + 3 y = 10 cos (x)

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{x} - 2 sin (x) + cos (x)

y = c_{1} e^{3 x} + c_{2} e^{x} - 2 sin (x) + cos (x)

y^{^{''}} + 4 y = 3 x (2 + cos (2 x))

y^{^{''}} - 2 y^{^{'}} + 5 y = 6 e^{t}, y (0) = 0

y^{^{''}} + 3 y^{^{'}} + 2 y = 10 u, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} - 2 y^{^{'}} - 2 = 0

\frac{d ^{2} y}{d x ^{2}} - y - 6 = 1

What is the solution for y^{''}-4y^'+3y=10cos(x) ?
The solution for y^{''}-4y^'+3y=10cos(x) is y=c_{1}e^{3x}+c_{2}e^x-2sin(x)+cos(x)