en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

t^2y^'+y=0

Solution

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

Solution

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

Solution steps

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

Solve separable ODE:

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

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

Popular Examples

limit as x approaches 4 of x

x \to 4 lim (x)

inverse oflaplace (s+3)/((s+3)^2+4)inverse laplace

\frac{s + 3}{( s + 3 ) ^{2} + 4}

derivative of 1/2 cos(x/2)

\frac{d}{d x} (\frac{1}{2} cos (\frac{x}{2}))

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

derivative of (3x+2^6)

\frac{d}{d x} ((3 x + 2)^{6})

Frequently Asked Questions (FAQ)

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