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Popular >

y^'=b*y+c

Solution

y^{^{'}} = b \cdot y + c

Solution

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

Solution steps

y^{'} (t) = b \cdot y + c

Solve separable ODE:

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

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

Popular Examples

8x-5ysqrt(x^2+1)(dy)/(dx)=0,y(0)=-10

8 x - 5 y x^{2} + 1 \frac{d y}{d x} = 0, y (0) = - 10

(xtan(x))dx-(ycos(x))dy=0

(x tan (x)) d x - (y cos (x)) d y = 0

x^{^{'}} - a \cdot x = 0

(dy)/(dx)y^2-x^3=1

\frac{d y}{d x} y^{2} - x^{3} = 1

y^'=0.01y-100,y(0)=20000

y^{^{'}} = 0.01 y - 100, y (0) = 20000

Frequently Asked Questions (FAQ)

What is the solution for y^'=b*y+c ?
The solution for y^'=b*y+c is y=(e^{tb+c_{1}b})/b-c/b