Solving Homogenous First Order ODE Using Substitution

Question

Problem

Find the general solution of the following homogeneous equation.

$$ ty' = y + \sqrt { t^2 - y^2} $$

Attempt

I'm following the algorithm provided in section 1.5 of the William Adkins & Mark G. Davidson, Ordinary Differential Equations textbook on page 63, which suggests rearranging the equation into the form:

$$ y' = ... $$

where $ y = tv $ and $ y' = v + tv' $ (v is the substitution variable).

I haven't been able to get too far in the procedure to make the substitution:

$$ y' = \frac{y + \sqrt { t^2 - y^2}}{t} $$ $$ y' = \frac{\frac{1}{t}}{\frac{1}{t}} \frac{y + \sqrt { t^2 - y^2}}{t} $$ $$ y' = \frac{y}{t} + \frac{\sqrt { t^2 - y^2}}{t} $$

Notes

Can someone please help me reduce the RHS of the above equation to the form $ y' = f(\frac{y}{t}) $ (an equation in terms of $ (\frac{y}{t}) $), using the substitution $ y = tv $? For bonus points, please provide a solution to the ODE so myself and other viewers can check their answers!

Thanks in advance!

Solution

With the help of Isham, I was able to get the solution.

$$ ty' = y + \sqrt { t^2 - y^2 } $$

It is implied that $ t^2 - y^2 \geq 0 $, and so $ t^2 \geq y^2 \gt 0 $ or $ \lt 0 $.

$$ ty' - y = \sqrt { t^2 - y^2} $$ $$ \frac {1}{\sqrt {t^2} } (ty' - y) = \frac {1}{\sqrt {t^2} } \sqrt { t^2 - y^2} $$ $$ \frac {1}{t} (ty' - y) = \sqrt { \frac{t^2}{t^2} - \frac{y^2}{t^2}} $$ $$ y' - \frac{y}{t} = \sqrt { 1 - \frac{y^2}{t^2}} $$ $$ y' - \frac{y}{t} = \sqrt { 1 - (\frac{y}{t})^2} $$ $$ y' = \sqrt { 1 - (\frac{y}{t})^2} + \frac{y}{t} $$

Let $ v = \frac{y}{t} $. Then, $ y = tv $ and so $ y' = \frac{dy}{dt} = v + tv' $. Substituting for $ v $ in the above equation we get:

$$ v + tv' = \sqrt {1 - v^2} + v $$ $$ tv' = \sqrt {1 - v^2} $$

Note, since $ t^2 \geq y^2 $, $ \frac{y^2}{t^2} = v^2 \leq 1 $, so there are no equilibrium cases to consider.

$$ t\frac{dv}{dt} = \sqrt {1 - v^2} $$ $$ \frac{1}{\sqrt { 1 - v^2}}dv = \frac{1}{t}dt $$ $$ \int\frac{1}{\sqrt { 1 - v^2}}dv = \int\frac{1}{t}dt $$ $$ arcsin(v) = ln|t| + C_1, C_1 \in \mathbb R $$ $$ \frac {y}{t} = \sin{(ln|t| + C_1)} $$ $$ y = t \sin{(ln|t| + C_1)} $$

Answer 1

Your substitution is fine. $$ty' = y + \sqrt { t^2 - y^2}$$ $$ty' - y = \sqrt { t^2 - y^2}$$ $$\frac {ty' - y}{t^2} = \frac 1t \sqrt { 1 - (\frac yt)^2}$$ $$ \left (\frac yt \right )'= \frac 1t \sqrt { 1 - (\frac yt)^2}$$ $$ \int \frac {d\left (\frac yt \right )}{\sqrt { 1 - (\frac yt)^2}}= \int \frac {dt}t $$

Answer 2

Hint: To reduce, you need to substitute $y=vt$ in RHS and $y'=v+tv'$ in LHS. It would look like this:

$$v+tv'=\frac{tv+t\sqrt{1-v^2}}{t}$$

$$tv'=\sqrt{1-v^2}$$

Where $\displaystyle v'$ is $\displaystyle\frac{\mathrm{d}v}{\mathrm{d}t}$ Can you proceed?

Answer 3

According with the suggested

$$ y'=v+\sqrt{1-v^2} $$

with

$$ v = \frac yt $$

so

$$ y' = v + t v' $$

then

$$ v + t v' = v +\sqrt{1-v^2} $$

or

$$ \frac{dv}{\sqrt{1-v^2}} = \frac{dt}{t} $$