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I've been working on this question but just can't crack it. It is as follows.

Using $ \frac {dy}{dx}= \frac {a\sqrt{x^2+y^2}+by}{bx}$ where $ y(1)=0 $

and by letting $ v=\frac{y}{x} $ transform the above differential equation to an equation in $v$ and $x$. Find the implicit solution of the resulting separable equation. Then use the relationship $y=xv$ to obtain $y+\sqrt{x^2+y^2}=x^{\frac{a}{b}+1}$

I've tried separate the upper terms into two fractions which makes the second term simply $\frac {by}{bx}=v$ but I am still left with the nasty square root with the y on the inside and I can't shift it easily to solve as a first order DE. Also I was thinking about solving it as a linear first order ODE but we have a squared term. Sorry about formatting as well, first time posting a question.

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Just as you did $$\frac {dy}{dx}= \frac {a\sqrt{x^2+y^2}+by}{bx}$$ Let $$y=x v\implies y'=v+x v'$$ making the equation to be $$x v'=\frac ab \sqrt{1+v^2}$$ which is separable $$\int \frac{dx} x=\frac b a \int\frac{dv}{\sqrt{1+v^2}}$$ making $$\log(x)+C=\frac b a \sinh ^{-1}(v)$$ So $$v=\sinh\left(\frac ab \log(x)+C\right)\implies y=x \,\sinh\left(\frac ab \log(x)+C\right)$$

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