Submitted by haku on Tue, 07/10/2018 - 19:11
Forums:
Hello, I have a question concerning the following differential equation
$$x \frac{\mathrm dy}{\mathrm dx}=y+1$$
Separating variables and integrating:
$$\int \frac{1}{y+1} \mathrm dy=\int \frac 1x \mathrm dx$$
$$\ln|y+1|=\ln|x|+c$$
In the solutions to question 3a) in the Mixed Pure STEP I module, the following step is made:
$$y+1=Ax \,\,\,\,\, \text{(where} \ A=e^c).$$
Why is it possible to omit the modulus function?
Many thanks in advance.
Modulus
Let's keep it in for a minute:
\[
\ln |y+1| = \ln |x| + c \\
|y+1| = |x| \times \text{e}^c\\
|y+1| = k |x| \; \; \text{where }k=\text{e}^c
\]
$k$ is an unknown constant, but we do know that $k$ must be positive by the way we have set it up.
Now:
\[
|y+1| = k|x|\\
\implies (y+1) = \pm k (x)
\]
Now if we let $A$ be a constant which can be negative (so $A=\pm k$) we then have:
\[
y+1 = Ax
\]
So I suppose technically we should say that $A=\pm \text{e}^c$ to deal with the modulus signs.
Does this help?
Yes, thanks a lot!
Yes, thanks a lot!