What is the solution for y^{''}-2y^'-15=0 ?

The solution for y^{''}-2y^'-15=0 is y=c_{1}e^{2t}+c_{2}-(15t)/2

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y^{''}-2y^'-15=0

y^{^{''}} - 2 y^{^{'}} - 15 = 0

y = c_{1} e^{2 t} + c_{2} - \frac{15 t}{2}

Solution steps

y^{''} (t) - 2 y^{'} (t) - 15 = 0

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} - \frac{15 t}{2}

y = c_{1} e^{2 t} + c_{2} - \frac{15 t}{2}

(D^{2} - 3 D + 2) y = x e^{x}

y^{^{''}} + y = 8 sin (x) + 4 cos (x) + 1

x^{^{''}} + 4 x^{^{'}} + 4 = 0

y^{^{''}} - 2 y^{^{'}} + y = \frac{e ^{t}}{( 1 + t ^{2} ) + 3 e ^{t}}

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

What is the solution for y^{''}-2y^'-15=0 ?
The solution for y^{''}-2y^'-15=0 is y=c_{1}e^{2t}+c_{2}-(15t)/2