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Popular >

y^'=1+ay

Solution

y^{^{'}} = 1 + a y

Solution

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

Solution steps

y^{'} (t) = 1 + a y (t)

Solve separable ODE:

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

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

Popular Examples

derivative of cos(5x^2-7x(10x-7))

\frac{d}{d x} (cos (5 x^{2} - 7 x) (10 x - 7))

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

integral of (-4x^{3/2}+3x^{2/3}-4x^2)

\int (- 4 x^{\frac{3}{2}} + 3 x^{\frac{2}{3}} - 4 x^{2}) d x

(\partial)/(\partial z)(sin(y))

\frac{\partial}{\partial z} (sin (y))

(dy)/(dx)=((y+x))/((y-x)),y(0)=1

\frac{d y}{d x} = \frac{( y + x )}{( y - x )}, y (0) = 1

Frequently Asked Questions (FAQ)

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