opepayflarpws

2022-06-22

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.

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.

assumintdz

Beginner2022-06-23Added 22 answers

Just as you did

$\frac{dy}{dx}=\frac{a\sqrt{{x}^{2}+{y}^{2}}+by}{bx}$

Let

$y=xv\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{y}^{\prime}=v+x{v}^{\prime}$

making the equation to be

$x{v}^{\prime}=\frac{a}{b}\sqrt{1+{v}^{2}}$

which is separable

$\int \frac{dx}{x}=\frac{b}{a}\int \frac{dv}{\sqrt{1+{v}^{2}}}$

making

$\mathrm{log}(x)+C=\frac{b}{a}{\mathrm{sinh}}^{-1}(v)$

So

$v=\mathrm{sinh}(\frac{a}{b}\mathrm{log}(x)+C)\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}y=x\phantom{\rule{thinmathspace}{0ex}}\mathrm{sinh}(\frac{a}{b}\mathrm{log}(x)+C)$

$\frac{dy}{dx}=\frac{a\sqrt{{x}^{2}+{y}^{2}}+by}{bx}$

Let

$y=xv\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{y}^{\prime}=v+x{v}^{\prime}$

making the equation to be

$x{v}^{\prime}=\frac{a}{b}\sqrt{1+{v}^{2}}$

which is separable

$\int \frac{dx}{x}=\frac{b}{a}\int \frac{dv}{\sqrt{1+{v}^{2}}}$

making

$\mathrm{log}(x)+C=\frac{b}{a}{\mathrm{sinh}}^{-1}(v)$

So

$v=\mathrm{sinh}(\frac{a}{b}\mathrm{log}(x)+C)\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}y=x\phantom{\rule{thinmathspace}{0ex}}\mathrm{sinh}(\frac{a}{b}\mathrm{log}(x)+C)$