How to solve $x'(t) =\frac {10 - x(t)}{2000}$, such that $x(0) = 0$, without using complex numbers

Question

Wolfram Alpha shows this intermediary step:

Solve the equation: $$c = 2000\, i π + 2000 \log(10)$$

It is however for a class where students have not yet learned complex numbers. How can they solve this?

Answer 1

This is just a first order differential equation. Try separation of variable or use integrating factor.

$$\frac{dx}{dt}+\frac{x}{2000}=\frac1{200}$$

Multiply by $\exp\left(\frac{t}{2000} \right)$,

$$\exp\left(\frac{t}{2000} \right)\frac{dx}{dt}+\exp\left(\frac{t}{2000} \right)\frac{x}{2000}=\frac1{200}\exp\left(\frac{t}{2000} \right)$$

it can be simplified to

$$\frac{d}{dt}\left(x \exp\left(\frac{t}{2000}\right)\right)=\frac1{200}\exp\left(\frac{t}{2000} \right)$$

Integrate both sides by $t$ and hopefully you can complete the rest.

Answer 2

This equation is separable $$\displaystyle \frac {dx}{dt} = -\frac {(x - 10)}{2000}, x(0) = 0$$ $$\int \frac {dx}{x-10}=-\frac 1 {2000}\int dt$$ $$\ln(x-10)=-\frac t {2000}+K$$ for $x<10$ $$ \displaystyle 10-x=Ke^{-\frac t {2000}}$$ $$ \displaystyle x=10+Ce^{-\frac t {2000}}$$ The constant for the initial condition given is $$x(0)=0 \implies C=-10$$ $$ \implies \displaystyle x(t)=10(1-e^{-\frac t {2000}})$$

Answer 3

Considering $$x' =\frac {10 - x}{2000}\implies x'+\frac { x}{2000}=\frac 1{200}$$ the homogeneous equation is simple $$x=c \,e^{-t/2000}$$ Using Now the veriation of parameters, the full equation becomes $$c' \,e^{-t/2000}=\frac{1}{200}\implies c'=\frac{e^{t/2000} }{200}\implies c=10 e^{t/2000}+K$$ Back to $x$ $$x=K e^{-t/2000}+10$$ So, if $x(0)=0$, then $K=-10$.

Answer 4

Let $y=10-x$. The equation becomes $-y'=ya$ wifh $a=1/2000$ and $y(0)=10$.

This is $y'/y=-a$ so $\ln y=-at+c$ or $y=Ce^{-at}$ or $x=10-Ce^{-at}$.

Set $t=0$ to find $C=10$.