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Popular >

x^'=b^2+x

Solution

x^{^{'}} = b^{2} + x

Solution

x = e^{t + c_{1}} - b^{2}

Solution steps

x^{'} (t) = b^{2} + x

Solve separable ODE:

x = e^{t + c_{1}} - b^{2}

x = e^{t + c_{1}} - b^{2}

Graph

Popular Examples

(dy)/(dx)=0.04-(3y)/(880)

\frac{d y}{d x} = 0.04 - \frac{3 y}{880}

(dy)/(dx)y=12(3x+1)^{2/3}(4x-2)^{5/4}

\frac{d y}{d x} y = 12 (3 x + 1)^{\frac{2}{3}} (4 x - 2)^{\frac{5}{4}}

y^{^{'}} = 1 - 4 y

(dv)/(dx)=-0.002v-11

\frac{d v}{d x} = - 0.002 v - 11

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

\frac{d y}{d x} = (3 x^{2} y), y (1) = 1

Frequently Asked Questions (FAQ)

What is the solution for x^'=b^2+x ?
The solution for x^'=b^2+x is x=e^{t+c_{1}}-b^2