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Popular >

y^'-ay+b=0

Solution

y^{^{'}} - a y + b = 0

Solution

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

Solution steps

y^{'} (t) - a y (t) + b = 0

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 6 to 8 of (56)/((x-6)^3)

\int_{6}^{8} \frac{56}{( x - 6 ) ^{3}} d x

(\partial)/(\partial x)((x^3+y^2)/(x-y))

\frac{\partial}{\partial x} (\frac{x ^{3} + y ^{2}}{x - y})

derivative of (3x+5^4)

\frac{d}{d x} ((3 x + 5)^{4})

integral from 1 to 2 of (cos(2x)-2/x)

\int_{1}^{2} (cos (2 x) - \frac{2}{x}) d x

derivative of f(x)=(-4x^2+9)^2 derivative of

f (x) = (- 4 x^{2} + 9)^{2}

Frequently Asked Questions (FAQ)

What is the solution for y^'-ay+b=0 ?
The solution for y^'-ay+b=0 is y=(e^{ta+c_{1}a})/a+b/a