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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

tangent of f(x)=2x^2+x-1,\at x=-2 tangent of

f (x) = 2 x^{2} + x - 1, at x = - 2

derivative of xe^{(-(x^2/8)})

\frac{d}{d x} (x e^{(- \frac{x ^{2}}{8})})

derivative of f(x)=-1/6 (x^{-6}-x^{12})derivative of

f (x) = - \frac{1}{6} (x^{- 6} - x^{12})

xy^'+y=x^{3/2}y^{5/2}ln(x)

x y^{^{'}} + y = x^{\frac{3}{2}} y^{\frac{5}{2}} ln (x)

tangent of e^{6x}tangent of

e^{6 x}

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