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Popular >

-x^'=bx

Solution

- x^{^{'}} = b x

Solution

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

Solution steps

- x^{'} (t) = b x (t)

Solve separable ODE:

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

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

Graph

Popular Examples

y^'=2 x/((y+x^2y)),y(0)=-2

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

(dy)/(dx)y=x^{4x}

\frac{d y}{d x} y = x^{4 x}

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

\frac{d y}{d x} y = cosh^{2} (3 x) - sinh (x^{2})

(dy)/(dx)=(3+12x)/(xy^2),x>0

\frac{d y}{d x} = \frac{3 + 12 x}{x y ^{2}}, x > 0

y^'-((t+1)y)/(t^2)= 1/t+1/(t^2)

y^{^{'}} - \frac{( t + 1 ) y}{t ^{2}} = \frac{1}{t} + \frac{1}{t ^{2}}

Frequently Asked Questions (FAQ)

What is the solution for -x^'=bx ?
The solution for -x^'=bx is x=e^{-bt+c_{1}}