What is the solution for y^{''}-25y=0,y(0)=1,y^'(0)=5 ?

The solution for y^{''}-25y=0,y(0)=1,y^'(0)=5 is y=e^{5t}

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y^{''}-25y=0,y(0)=1,y^'(0)=5

y^{^{''}} - 25 y = 0, y (0) = 1, y^{^{'}} (0) = 5

y = e^{5 t}

Solution steps

y^{''} (t) - 25 y = 0

Solve linear ODE:

y = e^{5 t}

y = e^{5 t}

y^{^{''}} + 6 y^{^{'}} + 34 y = 0, y (0) = 2, y^{^{'}} (0) = - 5

9 y^{^{''}} + 16 y = 0

a \frac{d ^{2} y}{d x ^{2}} + b \frac{d y}{d x} + cy = 0

2 y^{^{''}} + 16 y = 0

\frac{d ^{2} x}{d t ^{2}} + 3 \frac{d x}{d t} + 6 x = 0

What is the solution for y^{''}-25y=0,y(0)=1,y^'(0)=5 ?
The solution for y^{''}-25y=0,y(0)=1,y^'(0)=5 is y=e^{5t}