en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

t^2y^'+y^2=ty

Solution

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

Solution

y = \frac{t}{ln ( t ) + c _{1}}

Solution steps

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

Solve Bernoulli ODE:

y = \frac{t}{ln ( t ) + c _{1}}

y = \frac{t}{ln ( t ) + c _{1}}

Popular Examples

derivative of (3x^2+x)(5x^3+1)derivative of

(3 x^{2} + x) (5 x^{3} + 1)

integral of (-1)/(e^x)

\int \frac{- 1}{e ^{x}} d x

integral of x(x^4+1/4 x)

\int x (x^{4} + \frac{1}{4} x) d x

integral of y^3e^{y^4}

\int y^{3} e^{y^{4}} d y

derivative of cos((pix^9))

\frac{d}{d x} (cos ((π x)^{9}))

Frequently Asked Questions (FAQ)

What is the solution for t^2y^'+y^2=ty ?
The solution for t^2y^'+y^2=ty is y= t/(ln(t)+c_{1)}