What is the solution for y^{''}+3y^'-10y=0,y(0)=α,y^'(0)=5 ?

The solution for y^{''}+3y^'-10y=0,y(0)=α,y^'(0)=5 is y=e^{2t}(α+(5-2α)/7)-(e^{-5t}(5-2α))/7

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y^{''}+3y^'-10y=0,y(0)=α,y^'(0)=5

y^{^{''}} + 3 y^{^{'}} - 10 y = 0, y (0) = α, y^{^{'}} (0) = 5

y = e^{2 t} (α + \frac{5 - 2 α}{7}) - \frac{e ^{- 5 t} ( 5 - 2 α )}{7}

Solution steps

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

Solve linear ODE:

y = e^{2 t} (α + \frac{5 - 2 α}{7}) - \frac{e ^{- 5 t} ( 5 - 2 α )}{7}

y = e^{2 t} (α + \frac{5 - 2 α}{7}) - \frac{e ^{- 5 t} ( 5 - 2 α )}{7}

y^{^{''}} + 49 y = 0

x^{^{''}} + a x^{^{'}} + 2^{2} x = 0

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

\frac{d ^{2} x}{d t ^{2}} + a \cdot \frac{d x}{d t} = 0

\frac{d ^{2} y}{d x ^{2}} + \frac{d y}{d x} + 4 y = 0

What is the solution for y^{''}+3y^'-10y=0,y(0)=α,y^'(0)=5 ?
The solution for y^{''}+3y^'-10y=0,y(0)=α,y^'(0)=5 is y=e^{2t}(α+(5-2α)/7)-(e^{-5t}(5-2α))/7