Question

[Question from a DE textbook]
\[\boxed{
\newcommand \dd [1] {\,\mathrm d#1} % infinitesimal
\newcommand \de [2] {\frac{\mathrm d #1}{\mathrm d#2}} % first order derivative
\newcommand \ps [1] {\left(#1\right)} % dynamic parentheses
}\]
Solve the following differential equation:\[x\de yx = y+\sqrt{x^2+y^2}\]
Answer provided in the back of the book:\[y(x) = \frac12\ps{Ax^2-\frac1A}\]
Answer from wolfram-alpha:\[y(x) = x\sinh\ps{\ln kx}\]
Answer from my calculation (below):\[y(x) = kx^2-\sqrt{x^2+y^2}\]

Why does my answer not match with either of the other answers?
Have I made mistakes?

Answer 1

\begin{align}
x\de yx
&= y+\sqrt{x^2+y^2} \\
\de yx
&= y/x +\sqrt{1+(y/x)^2} \\[-2ex]
&&\text{let } y/x &= v \\
&& y &= vx \\
&& \de yx &= v+x\de vx \\[-2ex]
v+x\de vx
&= v +\sqrt{1+v^2} \\
x\de vx
&= \sqrt{1+v^2} \\
\frac{\dd v}{\sqrt{1+v^2}}
&= \frac{\dd x}x \\
\int\frac{\dd v}{\sqrt{1+v^2}}
&= \int\frac{\dd x}x \\
\ln\ps{v+\sqrt{1+v^2}}
&= \ln kx \\
v+\sqrt{1+v^2}
&= kx \\
y/x+\sqrt{1+(y/x)^2}
&= kx \\
y+\sqrt{x^2+y^2}
&= kx^2 \\
y(x)&= kx^2-\sqrt{x^2+y^2}
\end{align}

Answer 2

With a definition of the hyperbolic sine:\[\sinh \theta=\frac{e^\theta-e^{-\theta}}2\]
The wolfram-alpha result:
\[y(x)=x\sinh(\ln kx)\\
\qquad= x\ps{\frac{e^{\ln kx}-e^{-\ln kx}}2}\\
\qquad= x\ps{\frac{kx-\frac1{kx}}2}\\
\qquad= \frac12\ps{kx^2-\frac1k}
\]
can be show to equivalent the answer in the back of the book.

But I don't know how to get my answer to be equilivalent

Answer 3

your answer is equivalent to other answers,
just isolate y :)

Answer 4

(y-kx^2)^2 = x^2 + y^2

y^2 gets cancelled :)

Answer 5

This is such a dirty thing to put, a logarithm inside a hyperbolic trig function?

\[\sinh (\ln a) = \frac{e^{\ln a}-e^{-\ln a}}{2} = \frac{1}{2}a + \frac{-1}{2} a^{-1}\]
It's sorta awkward but maybe that's just a personal preference.

Answer 6

Got it now, thank you @hartnn !

\begin{align}
y+\sqrt{x^2+y^2}
&= kx^2 \\
y-kx^2
&= \sqrt{x^2+y^2} \\
y^2-2kyx^2+k^2x^4
&= x^2+y^2 \\
-2kyx^2+k^2x^4
&= x^2 \\
-2ky+k^2x^2
&= 1 \\
-2ky&= 1-k^2x^2 \\
y(x)&= \frac12\ps{kx^2-\frac1k}
\end{align}

Answer 7

As far as I can see, the only advantage of the hyperbolic-logarithm form is that the constant only appears once, (rather than twice).