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Popular >

x^'=a+b*x

Solution

x^{^{'}} = a + b \cdot x

Solution

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

Solution steps

x^{'} (t) = a + b \cdot x

Solve separable ODE:

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

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

Popular Examples

sin(x)y^'+(sin^2(x))/(cos(x))y=0

sin (x) y^{^{'}} + \frac{sin ^{2} ( x )}{cos ( x )} y = 0

y^'=-1/5 y^2,y(0)=0.25

y^{^{'}} = - \frac{1}{5} y^{2}, y (0) = 0.25

4(dy)/(dx)=(e^ysin^2(x))/(ysec(x))

4 \frac{d y}{d x} = \frac{e ^{y} sin ^{2} ( x )}{y sec ( x )}

(dy)/(dx)+x*y=x*y^2

\frac{d y}{d x} + x \cdot y = x \cdot y^{2}

(dy)/(dx)=\sqrt[3]{y},y(0)= 52/100

Frequently Asked Questions (FAQ)

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