What is the solution for y^{''}-2y^'-15=0,y(0)=2,y^'(0)=6 ?

The solution for y^{''}-2y^'-15=0,y(0)=2,y^'(0)=6 is y=(2+19/4)e^{2t}-19/4-(15t)/2

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y^{''}-2y^'-15=0,y(0)=2,y^'(0)=6

y^{^{''}} - 2 y^{^{'}} - 15 = 0, y (0) = 2, y^{^{'}} (0) = 6

y = (2 + \frac{19}{4}) e^{2 t} - \frac{19}{4} - \frac{15 t}{2}

Solution steps

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

Solve linear ODE:

y = (2 + \frac{19}{4}) e^{2 t} - \frac{19}{4} - \frac{15 t}{2}

y = (2 + \frac{19}{4}) e^{2 t} - \frac{19}{4} - \frac{15 t}{2}

y^{^{''}} - 5 y^{^{'}} + 6 y = e^{3 x} - x^{2}

\frac{d ^{2} y}{d x ^{2}} + 4 y = - 3 x^{2} + 2 x + 3

y^{^{''}} + 16 y = g (t), y (0) = 5, y^{^{'}} (0) = 0

y^{^{''}} - 6 y^{^{'}} + 9 y = e^{3 t} ln (t)

y^{^{''}} - y^{^{'}} - 2 y = 1, y (0) = 1, y^{^{'}} (0) = 3

What is the solution for y^{''}-2y^'-15=0,y(0)=2,y^'(0)=6 ?
The solution for y^{''}-2y^'-15=0,y(0)=2,y^'(0)=6 is y=(2+19/4)e^{2t}-19/4-(15t)/2