en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'=e^{4x-y}

Solution

y^{^{'}} = e^{4 x - y}

Solution

y = ln (\frac{1}{4} e^{4 x} + c_{1})

Solution steps

y^{'} (x) = e^{4 x - y}

Solve separable ODE:

y = ln (\frac{1}{4} e^{4 x} + c_{1})

y = ln (\frac{1}{4} e^{4 x} + c_{1})

Graph

Popular Examples

(1+y)*dx+((dy)/((x^2-2*x)))=0

(1 + y) \cdot d x + (\frac{d y}{( x ^{2} - 2 \cdot x )}) = 0

(dy)/(dt)=y^2-9

\frac{d y}{d t} = y^{2} - 9

derivative of ax^2y= derivative of 7

\frac{d}{d x} (a x^{2} y) = \frac{d}{d x} (7)

4/x (dy)/(dx)= 1/(y^3)

\frac{4}{x} \frac{d y}{d x} = \frac{1}{y ^{3}}

(1-u)udw+(1-u)wdu=(1+u)dw

(1 - u) u d w + (1 - u) w d u = (1 + u) d w

Frequently Asked Questions (FAQ)

What is the solution for y^'=e^{4x-y} ?
The solution for y^'=e^{4x-y} is y=ln(1/4 e^{4x}+c_{1})