Why is this equation non separable?

[brunette15]

Hi everyone,
I am trying to find any particular solution for the equation dy/dx + y = 1.
I have been told it is not separable.

I have done the following:
dy/dx = 1-y
integral of 1/(1-y) dy = integral
-loge(1-y) = c
e^-c = 1-y
y = 1- e^-c
let c = 0
y = 1-1
A particular solution is y= 0.

My question is why would my questions be mathematically incorrect?

 

[MarkFL]

I've moved this thread to our Differential Equations forum as that's a better fit for the question being asked. :D

The equation you posted is in fact separable, but I find it much easier to solve if we treat it as a linear equation, and multiply through by $e^x$ to obtain:

$$e^x\d{y}{x}+ye^x=e^x$$

Rewrite the left side, observing it is the differentiation of a product:

$$\frac{d}{dx}\left(e^xy\right)=e^x$$

Now, integrate through with respect to $x$:

$$e^xy=e^x+C$$

Divide through now by $e^x$ to obtain the explicit solution:

$$y(x)=1+Ce^{-x}$$

Notice that for suitable choice of $C$, we get the trivial solution $y\equiv1$ that you would technically lose by treating it as a separable equation.

 

[I like Serena]

dy/dx = 1-y
integral of 1/(1-y) dy = integral
-loge(1-y) = c

Hey brunette55, (Smile)

That should be $-\ln(1-y)=x+c$.
Do you see why?

 

[Prove It]

Hi everyone,
I am trying to find any particular solution for the equation dy/dx + y = 1.
I have been told it is not separable.

I have done the following:
dy/dx = 1-y
integral of 1/(1-y) dy = integral ?

Your approach is valid, this equation is separable as you have shown. Your problem is that you did not integrate the RHS with respect to x. This is why I much prefer doing every step...

$\displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} &= 1 - y \\ \frac{1}{1 - y}\,\frac{\mathrm{d}y}{\mathrm{d}x} &= 1 \\ \int{ \frac{1}{1 - y }\,\frac{\mathrm{d}y}{\mathrm{d}x} \, \mathrm{d}x} &= \int{ 1 \,\mathrm{d}x} \\ \int{ \frac{1}{1 - y}\,\mathrm{d}y} &= x + C_1 \\ -\ln{ \left| 1 - y \right| } + C_2 &= x + C_1 \end{align*}$

Can you continue?

 

[brunette15]

This all clears up a lot, thanks everyone! :)