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Popular >

25y^{''}-y=0

Solution

25 y^{^{''}} - y = 0

Solution

y = c_{1} e^{\frac{t}{5}} + c_{2} e^{- \frac{t}{5}}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{\frac{t}{5}} + c_{2} e^{- \frac{t}{5}}

y = c_{1} e^{\frac{t}{5}} + c_{2} e^{- \frac{t}{5}}

Popular Examples

(d^2y)/(dx^2)+4(dy)/(dx)+13y=0

\frac{d ^{2} y}{d x ^{2}} + 4 \frac{d y}{d x} + 13 y = 0

8y^{''}+12y^'+5y=0

8 y^{^{''}} + 12 y^{^{'}} + 5 y = 0

6(d^2y)/(dx^2)+(dy)/(dx)-y=0

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

y^{''}+9y=0,y(0)=sin(3x)

y^{^{''}} + 9 y = 0, y (0) = sin (3 x)

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

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

Frequently Asked Questions (FAQ)

What is the solution for 25y^{''}-y=0 ?
The solution for 25y^{''}-y=0 is y=c_{1}e^{t/5}+c_{2}e^{-t/5}