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Popular >

y^'-by=0

Solution

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

Solution

y = e^{b t + c_{1}}

Solution steps

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

Solve separable ODE:

y = e^{b t + c_{1}}

y = e^{b t + c_{1}}

Popular Examples

(dy)/(dx)=y^2e^xsin(2x),y(0)=1

\frac{d y}{d x} = y^{2} e^{x} sin (2 x), y (0) = 1

x^'=(x)(1-x^{-10})

x^{^{'}} = (x) (1 - x^{- 10})

y^{^{'}} = y sin (3 x + 1)

y^'=(sin(x)e^{x+2y})/y

y^{^{'}} = \frac{sin ( x ) e ^{x + 2 y}}{y}

(dN)/(dt)= 3/2 N

\frac{d N}{d t} = \frac{3}{2} N

Frequently Asked Questions (FAQ)

What is the solution for y^'-by=0 ?
The solution for y^'-by=0 is y=e^{bt+c_{1}}