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Popular >

t^2y^'+5t^2y=0

Solution

t^{2} y^{^{'}} + 5 t^{2} y = 0

Solution

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

Solution steps

t^{2} y^{'} (t) + 5 t^{2} y = 0

Solve separable ODE:

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

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

Popular Examples

(2 + x) y^{^{'}} = 2 y

(dx)/(dt)-0.04x=5000,x(0)=18000

\frac{d x}{d t} - 0.04 x = 5000, x (0) = 18000

y^'=(y^2)/(x^2y+xy)

y^{^{'}} = \frac{y ^{2}}{x ^{2} y + x y}

(dy)/(dx)y=x(2x+1)^{1/3}

\frac{d y}{d x} y = x (2 x + 1)^{\frac{1}{3}}

(dy}{dx}=\frac{ln(2))/7 y-80000

\frac{d y}{d x} = \frac{ln ( 2 )}{7} y - 80000

Frequently Asked Questions (FAQ)

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