en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

5x^'+7x=15

Solution

5 x^{^{'}} + 7 x = 15

Solution

x = - \frac{e ^{\frac{- 7 t - c _{1}}{5}}}{7} + \frac{15}{7}

Solution steps

5 x^{'} (t) + 7 x = 15

Solve separable ODE:

x = - \frac{e ^{\frac{- 7 t - c _{1}}{5}}}{7} + \frac{15}{7}

x = - \frac{e ^{\frac{- 7 t - c _{1}}{5}}}{7} + \frac{15}{7}

Popular Examples

(dy)/(dx)=yx^2-1.1y,y(0)=1

\frac{d y}{d x} = y x^{2} - 1.1 y, y (0) = 1

y^{^{'}} + 2 y = - 4

y^'=-(5x^4-5)/(4y)

y^{^{'}} = - \frac{5 x ^{4} - 5}{4 y}

(e^4-y)cos(x)(dy)/(dx)=e^4sin(x)

(e^{4} - y) cos (x) \frac{d y}{d x} = e^{4} sin (x)

y^'=y-2y^2,y(0)=t(t)

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

Frequently Asked Questions (FAQ)

What is the solution for 5x^'+7x=15 ?
The solution for 5x^'+7x=15 is x=-(e^{(-7t-c_{1})/5})/(7)+15/7