# Cannot solve the DE

#### idontknow

How to solve the equation ? the solution in the book is $$\displaystyle y=ex$$.

$$\displaystyle y'=e^{xy'/y}$$.

#### skipjack

Forum Staff
$y = ce^{x/c}$, where $c$ is a non-zero constant.

The graphs of the above solutions have a common tangent line with equation $y = ex$, corresponding to a singular solution of the differential equation.

#### idontknow

My approach after hours : set $$\displaystyle x=e^{t}$$ ; $$\displaystyle \; y'=\frac{dy}{dt}\cdot \frac{dt}{dx}=e^{e^t dy/ydx}$$.

$$\displaystyle e^{-t} dy/dt =e^{dy/ydt }$$ ; $$\displaystyle \; dy/dt=e^{dy/ydt +t } \;$$ ; $$\displaystyle \; \ln dy =\ln dt + \ln e^t +\ln e^{dy/ydt}=\ln e^t dt +\ln e^{dy/ydt}$$.

$$\displaystyle \ln \frac{dy}{dy/ydt} =\ln e^t dt \;$$ ; $$\displaystyle \; yd(C_1 t)=e^t d(C_2 t)$$.

$$\displaystyle y=\frac{C_2 }{C_1 } e^t =ce^t=cx$$.

$$\displaystyle (cx)'=e^{x(cx)'/cx}=e$$.
$$\displaystyle c=e$$ and the general solution: $$\displaystyle y=ex$$.

#### skipjack

Forum Staff
$$\displaystyle \ln \frac{dy}{dy/ydt} =\ln e^t dt$$
That should be $$\displaystyle \ln \frac{dy}{e^{dy/ydt}} =\ln e^t dt \,$$.

You can't obtain the book's solution as the general solution, because it's a singular solution.

idontknow

#### idontknow

That should be $$\displaystyle \ln \frac{dy}{e^{dy/ydt}} =\ln e^t dt \,$$.
Yes , my mistake , I will post the correct version later.

#### idontknow

$$\displaystyle y'$$ must be a power of e. we can set $$\displaystyle y=e^{\phi (x)}$$.

$$\displaystyle (e^{\phi})'=e^{\frac{x(e^{\phi})'}{y}}\;$$ ; $$\displaystyle \; e^{\phi} \phi ' =e^{xe^{\phi}\phi ' /e^{\phi} }=e^{x\phi '}$$.

$$\displaystyle \ln e^{\phi} +\ln \phi ' =x\phi ' \;$$ ; $$\displaystyle \; \ln \phi ' =x\phi ' -\phi \;$$ ; $$\displaystyle \; \frac{d}{dx} \ln \phi ' = \frac{d}{dx} [x\phi '-\phi]$$.

$$\displaystyle \frac{\phi ''}{\phi '} =x\phi '' +\phi ' -\phi ' =x\phi '' \;$$ ; $$\displaystyle \; \phi_1 '' [x-\frac{1}{\phi_2 '}]=0$$. (we need only $$\displaystyle \phi _1$$)

$$\displaystyle \phi _1 '' =0 \: \implies \: \phi_1=\ln y = Cx \: \implies \: y=e^{Cx}=cx$$.

$$\displaystyle (cx)'=e^{x(cx)'/cx}=e \: \implies \: y=ex .$$
The book does not mention about the general solution , just y=ex.

#### skipjack

Forum Staff
$$\displaystyle \phi _1 '' =0 \: \implies \: \phi_1=\ln y = Cx \: \implies \: y=e^{Cx}=cx$$.
That's not quite right. If you correct it, you can get the general solution I gave.

Solving for $\phi_2$ leads to the singular solution $y = ex$.

idontknow

#### idontknow

$$\displaystyle \ln \phi ' =x\phi ' -\phi$$ is known as Clairaut's DE : $$\displaystyle \: s(t)=ts'(t)+f(s'(t))$$.

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