en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'-2ty=0

Solution

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

Solution

y = e^{t^{2} + c_{1}}

Solution steps

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

Solve separable ODE:

y = e^{t^{2} + c_{1}}

y = e^{t^{2} + c_{1}}

Popular Examples

sin(y)cos(x)dy=cos(y)sin(x)dx

sin (y) cos (x) d y = cos (y) sin (x) d x

(dy)/(dx)= 1/5 sqrt(y)cos^2(sqrt(y))

\frac{d y}{d x} = \frac{1}{5} y cos^{2} (y)

(dp)/(dx)=k(a-p)

\frac{d p}{d x} = k (a - p)

(x^2-x)y^'-(y^2+y)=0

(x^{2} - x) y^{^{'}} - (y^{2} + y) = 0

(dy)/(dx)=yx^3-xy

\frac{d y}{d x} = y x^{3} - x y

Frequently Asked Questions (FAQ)

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