en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'=1-x+4y

Solution

y^{^{'}} = 1 - x + 4 y

Solution

y = \frac{x}{4} - \frac{3}{16} + c_{1} e^{4 x}

Solution steps

y^{'} (x) = 1 - x + 4 y

Solve linear ODE:

y = \frac{x}{4} - \frac{3}{16} + c_{1} e^{4 x}

y = \frac{x}{4} - \frac{3}{16} + c_{1} e^{4 x}

Popular Examples

(\partial)/(\partial x)(5e^{x^6y})

\frac{\partial}{\partial x} (5 e^{x^{6} y})

simplify (3x+6)/(2sqrt(x+3))simplify

\frac{3 x + 6}{2 x + 3}

integral from-4 to 2 of-(y^2)/2-y+4

\int_{- 4}^{2} - \frac{y ^{2}}{2} - y + 4 d y

derivative of f(x)=(x^5)/7 derivative of

f (x) = \frac{x ^{5}}{7}

integral of sqrt(cos(2x-2))

\int cos (2 x - 2) d x

Frequently Asked Questions (FAQ)

What is the solution for y^'=1-x+4y ?
The solution for y^'=1-x+4y is y= x/4-3/16+c_{1}e^{4x}