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Popular >

x^'+ax=b

Solution

x^{^{'}} + a x = b

Solution

x = - \frac{e ^{- a t - c_{1} a}}{a} + \frac{b}{a}

Solution steps

x^{'} (t) + a x (t) = b

Solve separable ODE:

x = - \frac{e ^{- a t - c_{1} a}}{a} + \frac{b}{a}

x = - \frac{e ^{- a t - c_{1} a}}{a} + \frac{b}{a}

Popular Examples

x^{^{'}} = 1 - 2.2 x

y^{^{'}} = 4 x^{3} y^{4}

(dy)/(dx)=6x^4sqrt(y),y>0

\frac{d y}{d x} = 6 x^{4} y, y > 0

(dy)/(dx)-2y^2cos^2(x)=3x^2y^2

\frac{d y}{d x} - 2 y^{2} cos^{2} (x) = 3 x^{2} y^{2}

(dy)/(dx)y=3x^5+4x^2+x^2+1000

\frac{d y}{d x} y = 3 x^{5} + 4 x^{2} + x^{2} + 1000

Frequently Asked Questions (FAQ)

What is the solution for x^'+ax=b ?
The solution for x^'+ax=b is x=-(e^{-at-c_{1}a})/a+b/a