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Popular >

(dx)/(dt)=a-bx^2

Solution

\frac{d x}{d t} = a - b x^{2}

Solution

x = - \frac{a e ^{2 a b t + 2 c_{1} a b}}{b ( 1 - e ^{2 a b (t + c_{1})} )} - \frac{a}{b ( 1 - e ^{2 a b (t + c_{1})} )}

Solution steps

\frac{d x}{d t} = a - b x (t)^{2}

Solve separable ODE:

x = - \frac{a e ^{2 a b t + 2 c_{1} a b}}{b ( 1 - e ^{2 a b (t + c_{1})} )} - \frac{a}{b ( 1 - e ^{2 a b (t + c_{1})} )}

x = - \frac{a e ^{2 a b t + 2 c_{1} a b}}{b ( 1 - e ^{2 a b (t + c_{1})} )} - \frac{a}{b ( 1 - e ^{2 a b (t + c_{1})} )}

Popular Examples

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

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

(dV)/(dt)=60-3/100 V

\frac{d V}{d t} = 60 - \frac{3}{100} V

y^'=(xy^2)/(sqrt(1+x^2)),y(0)=1

y^{^{'}} = \frac{x y ^{2}}{1 + x ^{2}}, y (0) = 1

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

\frac{d y}{d x} = \frac{( y ^{2} + y )}{( x ^{2} - x )}

(dy)/(dx)+20xy=0

\frac{d y}{d x} + 20 x y = 0

Frequently Asked Questions (FAQ)

What is the solution for (dx)/(dt)=a-bx^2 ?
The solution for (dx)/(dt)=a-bx^2 is