What is the solution for y^{''}-3y^'-10=0,y(0)=1,y^'(0)=10 ?

The solution for y^{''}-3y^'-10=0,y(0)=1,y^'(0)=10 is y=(1+31/9)e^{3t}-31/9-(10t)/3

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y^{''}-3y^'-10=0,y(0)=1,y^'(0)=10

y^{^{''}} - 3 y^{^{'}} - 10 = 0, y (0) = 1, y^{^{'}} (0) = 10

y = (1 + \frac{31}{9}) e^{3 t} - \frac{31}{9} - \frac{10 t}{3}

Solution steps

y^{''} (t) - 3 y^{'} (t) - 10 = 0

Solve linear ODE:

y = (1 + \frac{31}{9}) e^{3 t} - \frac{31}{9} - \frac{10 t}{3}

y = (1 + \frac{31}{9}) e^{3 t} - \frac{31}{9} - \frac{10 t}{3}

2 y^{^{''}} + 5 y^{^{'}} - y = 9

y^{^{''}} + 7 y^{^{'}} + 12 y = 340 cos (t)

y^{^{''}} + y = 2, + t y (0) = 1, y (0) = 1

y^{^{''}} + 25 y = 8 sec (5 x)

y^{^{''}} - y = (1 - e^{2 x})^{- \frac{1}{2}}

What is the solution for y^{''}-3y^'-10=0,y(0)=1,y^'(0)=10 ?
The solution for y^{''}-3y^'-10=0,y(0)=1,y^'(0)=10 is y=(1+31/9)e^{3t}-31/9-(10t)/3