Exercise: first-order linear differential equation

Question

This is a first-order linear differential equation:

$$y' = -ky + p$$

where $k$ and $p$ are constant. Based on my calculations, the solution is

$$y(x) = y(0) \cdot \exp^{-kx} + \; \dfrac{p}{k}$$

while my teacher's file says

$$y(x) = \dfrac{p}{k} + \left[y(0) - \dfrac{p}{k} \right] \exp^{-kx}$$

Which one is the right one?

Thank you in advance

Answer 1

The general solution is given by $$y(x)=\frac{p}{k}+Ce^{kx}$$ and now you will $$y(0)=\frac{p}{k}+C$$ to compute $$C$$

Answer 2

The solution given by your teacher is the right one. Assuming that the intial condition is given by $y(0)=y_0$ we get the following

\begin{align} y'&=-ky+p\\ y'&=-k\left(y-\frac pk\right)\\ \frac{y'}{y-\frac pk}&=-k\\ \int \frac{\mathrm dy}{y-\frac pk}&=-k\int\mathrm dx\\ \log\left(y-\frac pk\right)&=-kx+c\\ y-\frac pk&=ce^{-kx}\\ \end{align}

$$\therefore~y(x)~=~ce^{-kx}+\frac pk$$

Determining the constant $c$ by using $y(0)=y_0$ we get

$$y(0)=c+\frac pk=y_0\Rightarrow~c=y_0-\frac pk$$

$$\therefore~y(x)=\frac pk+\left[y_0-\frac pk\right]e^{-kx}$$

I cannot really tell where you went wrong but it is easy to check that your solution does not fulfill the initial condition since

$$y(x)=y_0e^{-kx}+\frac pk\Rightarrow y(0)=y_0+\frac pk\color{red}\neq y_0$$

Answer 3

Firstly, this is first-order.

We let $$y'+ky=p$$ Then use an integrating factor:

$$IF=e^{\int k dx}=e^{kx}$$

Then the trick: $$y\cdot IF =\int{IF\cdot RHS \ dx}$$ $$\to ye^{kx}=\int{pe^{kx}\ dx}$$ $$\to y=\frac pk +Ce^{-kx}$$

You appear correct therefore, unless the initial condition changes something.