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Separate DE via substitution

MacLaddy

Member
Jan 29, 2012
52
Hello all, it's been a long time. Hoping I can get some assistance with what is probably a simple substitution problem, yet it's flummoxing me.

$$\frac{dy}{dx} = \frac{y+t}{t}$$

I've tried substituting $$ v = y+t $$
$$ y = v - t $$
$$ \frac{dy}{dx} = \frac{dv}{dt} - 1 $$
$$\frac{dv}{dt}-1 = \frac{v}{t}$$
That would be simple enough, but the -1 is throwing me off.
$$\frac{dv}{v}-\frac{1}{v} = \frac{dt}{t}$$
Definitely something wrong there...

Please show me the error in my ways?

Thanks much,
Mac
 

Chris L T521

Well-known member
Staff member
Jan 26, 2012
995
Hello all, it's been a long time. Hoping I can get some assistance with what is probably a simple substitution problem, yet it's flummoxing me.

$$\frac{dy}{dx} = \frac{y+t}{t}$$

I've tried substituting $$ v = y+t $$
$$ y = v - t $$
$$ \frac{dy}{dx} = \frac{dv}{dt} - 1 $$
$$\frac{dv}{dt}-1 = \frac{v}{t}$$
That would be simple enough, but the -1 is throwing me off.
$$\frac{dv}{v}-\frac{1}{v} = \frac{dt}{t}$$
Definitely something wrong there...

Please show me the error in my ways?

Thanks much,
Mac
For starters, the problem doesn't make sense if you have a $\dfrac{dy}{dx}$ there. Seeing your work makes me think that the original equation should be
\[\frac{dy}{dt}=\frac{y+t}{t}=\frac{y}{t}+1.\]
Here, you can make the substitution $y=ut\implies \dfrac{dy}{dt}=u+t\dfrac{du}{dt}$. Thus, you get that
\[u+t\frac{du}{dt}=\frac{ut}{t}+1\implies u+t\frac{du}{dt}=u+1\implies \frac{du}{dt}=\frac{1}{t}.\]
That last differential equation is simple enough to solve.

Can you take things from here?
 

MacLaddy

Member
Jan 29, 2012
52
For starters, the problem doesn't make sense if you have a $\dfrac{dy}{dx}$ there. Seeing your work makes me think that the original equation should be
\[\frac{dy}{dt}=\frac{y+t}{t}=\frac{y}{t}+1.\]
Here, you can make the substitution $y=ut\implies \dfrac{dy}{dt}=u+t\dfrac{du}{dt}$. Thus, you get that
\[u+t\frac{du}{dt}=\frac{ut}{t}+1\implies u+t\frac{du}{dt}=u+1\implies \frac{du}{dt}=\frac{1}{t}.\]
That last differential equation is simple enough to solve.

Can you take things from here?
Ahh, yes. $ \frac{dy}{dx} $ was a nearly-midnight-brain-... Well, it was mis-typed.


Piece of cake, thank you very much for your help. The book vaguely suggested doing it this way, but I didn't understand the way it was presented.

$$ \frac{du}{dt} = \frac{1}{t} $$
$$ \int{du} = \int{\frac{1}{t}dt} $$
$$ u = ln|t| + C $$
$$ \frac{y}{t} = ln|t|+C $$
$$ y(t) = tln|t| + tC $$

Thanks again,
Mac
 
Last edited:

Chris L T521

Well-known member
Staff member
Jan 26, 2012
995
Ahh, yes. $ \frac{dy}{dx} $ was a nearly-midnight-brain-... Well, it was mis-typed.


Piece of cake, thank you very much for your help. The book vaguely suggested doing it this way, but I didn't understand the way it was presented.

$$ \frac{du}{dt} = \frac{1}{t} $$
$$ \int{du} = \int{\frac{1}{t}dt} $$
$$ u = ln|t| + C $$
$$ \frac{y}{t} = ln|t|+C $$
$$ y(t) = tln|t| + tC $$

Thanks again,
Mac
Yea, some books don't explain it that well. However, $y=ut$ is one of the more common substitutions made when solving first order ODES, so I'd be sure to be familiar with/remember it! :)
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,191
The motivation for using Chris's substitution comes about from one of the standard forms:
\begin{align*}
\frac{dy}{dt} &= \frac{y+t}{t}\\
t \, dy&=(y+t) \, dt.
\end{align*}
From here, you can see that the differential equation is homogeneous, thus making Chris's substitution the preferred method of solution.

Another method would be the integrating factor, since the equation is first-order linear.