differential equation with linear coefficients other answer than in book

Question

I have a differential equation: $$ (x+2y-4)dx +(-2x+4y)dy = 0 $$

Since the coefficients of $dx$ and $dy$ are assumed to define line in the plain, so:

$$ 7y -3 = 0 $$ $$ 2x+1 = 0 $$

Point od the intersection of these lines is:$ (x,y) = (2,1) $

Next, I'm trying to move origin of the plain to the intesection point.

We know that relations between coordinates are:

$$ x = \bar{x} +2 $$ $$ y = \bar{y} +1 $$

where $\bar{x}$, $\bar{y}$ are coordinates measured from point $ (2,1)$

After substitution these relations, equation simplifices to form:

$$ (\bar{x} + 2 \bar{y})d\bar{x} + (-2\bar{x}+4\bar{y})d\bar{y} =0 $$

Now I'm trying to do substitution:

$$ \bar{x} = u \bar{y}, d\bar{x} = u d\bar{y} + \bar{y}du $$

So after substitution and simplifications I'm obtaining:

$$ (u^{2} + 4) \bar{y}d\bar{y} + (u+2)\bar{y}^2 du = 0$$

Now, I'm dividing equation both sides by $(u^2+4)(\bar{y}^2)$

and I'm obtaining equation with separable variables:

$$ \frac{d\bar{y}}{\bar{y}} + \frac{u+2}{u^2+4}du = 0 $$

Because: $$ \int \frac{u+2}{u^2+4} = arctan(\frac{u}{2}) + \frac{1}{2} ln|4+u^2| + C$$

So, I have a solution:

$$ ln|\bar{y}| + arctan(\frac{u}{2})+ \frac{1}{2}ln|(4+u^2)| = C $$

After substitution relationship between $\bar{x},x$ and $\bar{y},y$ and simplifications, I'm obtaining:

$$ ln|4(y-1)^{2} + (x-2)^2| + 2 arctan(\frac{x-2}{2y-2}) = C $$

Now I have a problem because answer from book to this exercise is:

$$ ln|4(y-1)^{2} + (x-2)^2| -2 arctan(\frac{2y-2}{x-2})+= C $$

I've check that this solutions we may obtain with substitution:

$$ \bar{y} = u\bar{x}, d\bar{y} = u d \bar{x} + \bar{x}du $$

Is my answer wrong? I would be grateful for explaining. Best regards

Answer 1

These two answers are equivalent. Since $\arctan\left(\frac{1}{a}\right)=\frac{\pi}{2}-\arctan(a)$ for positive $a$ and $\arctan\left(\frac{1}{a}\right)=-\frac{\pi}{2}-\arctan(a)$ for negative $a$, your answer and their answer only differ by a constant. So both answers are correct — you just need to note that the two constants $C$ are different from each other.

Answer 2

With the usual cautions about divide by zero for the solution with the initial conditions, the general solution is the same.

$\ln|4(y-1)^{2} + (x-2)^2| -2\arctan(\dfrac{2y-2}{x-2})= C$

$\ln|4(y-1)^{2} + (x-2)^2| +2\left(\pi/2-\arctan\left(\dfrac{2y-2}{x-2}\right)\right)= C+\pi$

$\ln|4(y-1)^{2} + (x-2)^2| +2\arctan\left(\dfrac{x-2}{2y-2}\right)= C+\pi$

$\ln|4(y-1)^{2} + (x-2)^2| +2\arctan\left(\dfrac{x-2}{2y-2}\right)= C_1$

Really $\arctan$ is a headhache all around the site. Not the best in formal rigor, but I always recommend this about $\text{atan2}$ to deal with $\arctan$ in an intuitive way.

Answer 3

If the point $(a,b)$ varies strictly on the inside of a quadrant, the term $\arctan(b/a)$ is a constant shift away from $\arg(a+ib)$. As such, $$ \arg(a+ib)=\arg(i·(a+ib))-\frac{\pi}2+k·2\pi=\arg(-b+ia)-\frac{\pi}2+k·2\pi $$ which then is only a constant shift away from $-\arctan(a/b)$.

Thus on all intervals where both of your expressions are defined, they only differ by a constant which means they describe the same solution family.