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Popular >

ax^'+bx=0,x(0)=e^{rt}

Solution

a x^{^{'}} + b x = 0, x (0) = e^{r (x, t) t}

Solution

a x^{'}

Solution steps

Find

a x^{^{'}} + b x \circ x (0)

given

a x^{^{'}} + b x = 0, x (0) (x, t) = e^{r (x, t) t}

Simplify

a x^{^{'}} + b x \circ x \cdot 0 : a x^{^{'}}

= a x^{^{'}}

Popular Examples

(4x^5-16x^3-2x)(1-y^2)-2y(x^2-4)y^'=0

(4 x^{5} - 16 x^{3} - 2 x) (1 - y^{2}) - 2 y (x^{2} - 4) y^{^{'}} = 0

derivative of x^2+derivative of y^2=1

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

y^'=(1+x)/(xy),y(1)=-4

y^{^{'}} = \frac{1 + x}{x y}, y (1) = - 4

(dy)/(dx)+4xy=4e^6x

\frac{d y}{d x} + 4 x y = 4 e^{6} x

(dv)/(dt)=0.4v+1000

\frac{d v}{d t} = 0.4 v + 1000

Frequently Asked Questions (FAQ)

What is the ax^'+bx=0,x(0)=e^{rt} ?
The ax^'+bx=0,x(0)=e^{rt} is ax^'