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Popular >

y^'=ay+b

Solution

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

Solution

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

Solution steps

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

Solve separable ODE:

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

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

Popular Examples

integral from 0 to 1/2 of 4/(1+4x^2)

\int_{0}^{\frac{1}{2}} \frac{4}{1 + 4 x ^{2}} d x

integral of 1/(2(x^2+1))

\int \frac{1}{2 ( x ^{2} + 1 )} d x

integral of 1/(x^2)x'

\int \frac{1}{x ^{2}} x^{'} d x

(5x-3)y^{''}+y^'=0

(5 x - 3) y^{^{''}} + y^{^{'}} = 0

inverse oflaplace 6 inverse laplace

6

Frequently Asked Questions (FAQ)

What is the solution for y^'=ay+b ?
The solution for y^'=ay+b is y=(e^{ta+c_{1}a})/a-b/a