# Differential equation change of variable form

For example if the differential equation is in the form
$$\frac{dy}{dx} = f(ax+by+c),$$ we put $ax+by+c=v$.

Please, help me. I am not able to solve these questions and please give me tips so that I can improve my mathematics :

1. $\frac{dy}{dx} + 1 = e^{x-y}$

2. $\frac{dy}{dx} = \sqrt{y-x}$

3. $\frac{dy}{dx} = \sin(x+y) + \cos(x+y)$

• Hints: to get you going. For $1.$, let $v = x - y$, so $v' = 1 - y'$. For $2.$, let $v = y - x$, so $v' = y' - 1$. See a pattern here? Substitute and solve and the substitute back. Now, try $3.$ yourself. – Moo Jul 21 '17 at 5:02

$$\frac{dy}{dx} = f(ax+by+c)$$ $ax+by+c=v \quad\to\quad y= \frac{1}{b} (v-ax-c) \quad\to\quad \frac{dy}{dx} =\frac{1}{b}\frac{dv}{dx}-\frac{a}{b}=f(v)$ $$\frac{dv}{dx}=a+bf(v) \qquad\text{is a separable ODE.}$$ $$dx=\frac{dv}{a+bf(v)}\quad\to\quad x=\int \frac{dv}{a+bf(v)} +c$$

We have to be aware that this method doesn't always allows to express $y(x)$ on a closed form.

• First, the above integral is not always solvable on a closed form for $x(v)$.

• Second, the inverse function of $x(v)$, say $v(x)=x^{-1}(v)$ cannot be always expressed on closed form.

Only in case of the two hurdles can be overcome, one get to $v(x)$ and $y(x)= \frac{1}{b} (v(x)-ax-c)$

FOR EXAMPLE :

1. $\frac{dy}{dx} + 1 = e^{x-y} \qquad v=x-y \quad\to\quad y(x)=\ln\left(C+\frac{1}{2}e^{2x}\right)$.

2. $\frac{dy}{dx} = \sqrt{y-x}\qquad v=y-x \quad\to\quad \frac{dv}{dx} =-1+ \sqrt{v}\quad\to\quad x=\int \frac{dx}{ -1+\sqrt{v}}+C$ $x=2\sqrt{v}+2\ln(-1+\sqrt{v})+c \quad$. The inverse function involves a special function, the Lambert W function. $\quad\to\quad y(x)=x+\left(W(Ce^{x/2}) +1\right)^2$.

3. $\frac{dy}{dx} = \sin(x+y) + \cos(x+y) \qquad v=x+y \quad\to\quad \frac{dv}{dx}= -1+\sin(v) + \cos(v)\quad\to\quad x=\int \frac{dv}{-1+\sin(v) + \cos(v)} +c = \ln(\sin(\frac{v}{2} )-\ln\left(\cos(\frac{v}{2})-\sin(\frac{v}{2})\right)+c \quad\to\quad v=2\cot^{-1}\left(1+Ce^{-x} \right)\quad\to\quad y(x)=-x+2\cot^{-1}\left(1+Ce^{-x} \right)$

In your first example, let
$$u=x-y$$ So we have,
$$\frac{du}{dx}=1-\frac{dy}{dx}$$
$$\frac{dy}{dx}=1-\frac{du}{dx}$$
Your equation reduces to
$$2-\frac{du}{dx}=e^u$$
$$\frac{du}{dx}=2-e^u$$
Which is variable separable form, also note that
$$\int\frac{dx}{a+be^{cx}}=\frac{1}{ac}\ln|\frac{e^{cx}}{a+be^{cx}}|+C$$

You'll get to know which substitution to use only from practice.