2
$\begingroup$

I am looking for a help with the below equation

$$\frac{\mathrm dy}{\mathrm dx}=\frac{x+y}{x}$$

I don't get the suggested answer $$y = x\ln(x) + c x$$

My attempt $$\mathrm dx(1+\frac{y}{x})=\mathrm dy$$ $$c+x+y\ln(x)=y$$ $$y=\frac{c+x}{1-\ln x}$$

$\endgroup$
2
  • 2
    $\begingroup$ Hint: Integrating Factor $\endgroup$
    – Moo
    Jul 17, 2016 at 3:44
  • $\begingroup$ The mistake in your derivation: Note that $y = y(x)$ is a function of $x$ so you cannot just integrate $dx\left(1+\frac{y}{x}\right)$ by holding $y$ constant (as you have done). $\endgroup$
    – Winther
    Jul 17, 2016 at 3:58

6 Answers 6

5
$\begingroup$

The problem with your approach is that:

$$ \int \left(1 + \frac{y}{x}\right) dx \neq C + x + y \ln x $$

Note that $y$ is a function of $x$, so you cannot treat it as a constant when you integrate.

One appropriate way to do this problem using an integrating factor:

Doing some algebra:

$$ y'(x) - y(x)/x = 1 $$

Let $\mu(x) = \exp(\int -\frac1x dx) = 1/x$. Multiplying both sides by $\mu$:

$$ y'(x)/x - y(x)/x^2 = 1/x $$

Note that $-1/x^2$ is the derivative of $1/x$:

$$ \frac{dy}{dx} \frac{1}{x} + \frac{d}{dx} \left( \frac{1}{x} \right) y(x) = \frac{1}{x} $$

Using the product rule:

$$ \frac{d}{dx} (y/x) = \frac{1}{x}$$

You can finish the rest yourself.

$\endgroup$
3
$\begingroup$

We have:

$$\frac{dy}{dx}=\frac{x+y}{x}=1+\frac{y}{x}$$

Suppose $y=xv(x)$ with $v(x):\mathbb{R}\rightarrow\mathbb{R}$

Then $$\frac{dy}{dx}=v+x\frac{dv}{dx}=1+v$$

Now notice we have:

$$x\frac{dv}{dx}=1$$ which is separable.

$$v(x)=\ln(x)+C,$$ where $C\in \mathbb{R}$

Thus $$y=x\ln(x)+Cx$$

$\endgroup$
2
$\begingroup$

Hint: $$\frac{dy}{dx}=\frac{x+y}{x}\implies xdy=(x+y)dx\implies xdy-ydx=xdx\implies\frac{xdy-ydx}{x^2}=\frac{dx}{x}$$

$\endgroup$
2
$\begingroup$

With $\mu(x)=e^{\int\frac{M_y-N_x}{N}dx}$, where $M=x$ and $N=-x-y$ then $\mu(x)=\frac{1}{x^2}$, Now we have: $\frac{1}{x^2}(xdy+(-x-ydx))=\frac{1}{x^2}*0\Rightarrow \frac{1}{x}dy+(-\frac{1}{x}-\frac{y}{x^2})dx=0 $, we take $f_y=\frac{1}{x}\Rightarrow f=\frac{y}{x}+h(x)$, after $f_x=-\frac{y}{x^2}+h'(x)=-\frac{1}{x}-\frac{y}{x^2}\Rightarrow h'(x)=-\frac{1}{x}$ then $h(x)=-\ln x$. The solution is $\frac{y}{x}-\ln x=c$ where $c$ is constant

$\endgroup$
2
$\begingroup$

Just another way to do it.

Considering $$\frac{dy}{dx}=\frac{x+y}{x}$$ define $y=z-x$ which leads to $$\frac{dz}{dx}-1=\frac{z}{x}$$ that is to say $$\frac{dz}{dx}-\frac{z}{x}=1$$ Solving the homogeneous equation leads to $z=C x$. Now, using the method of variation of parameters, this leads to $$C+x C'-C=1$$ that is to say $xC'=1$ then $C=\log(x)+K$ then $z=x(\log(x)+K)$ and finally $y=x(\log(x)+K)-x=x \log(x)+(K-1)x=x \log(x)+ Lx$.

$\endgroup$
1
$\begingroup$

Hint

$$xy'=x+y$$ $$xy'-y=x$$ This is Euler-Cauchy Equation,see

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.