$xy'=y\ln(xy)$ differential equations

Question

I am really struggling with this problem. This section of my book is about solutions of differential equations through substitutions. We were taught about three methods of substitution in this section. The first about Homogeneous equations. where you have something like $$ M(x,y) + N(x,y) = 0$$ and $$M(tx,ty) = t^\alpha *M(x,y)\; and\; N(tx,ty) = t^\alpha *N(x,y)$$ then using the substitution u=y/x or v=x/y

the other method we learned was Bernoullis equation where you have the DE in the form $$\frac{dy}{dx}+P(x)y = f(x)y^n$$

and the last we learned of was Reduction to separation of variables. Where you have $$\frac{dy}{dx}=f(Ax+By+C)$$ and you use subsitutions

The problem states determine an appropriate substitution and solve $$xy'=yln(xy)$$ I found an answer on yahoo answers but I am lost in the steps. https://answers.yahoo.com/question/index?qid=20150420091352AA9lvDl The answer given is the same as the correct answer. Can anyone explain the steps especially the part where it transitions from $$xydu=xdy+ydx = y(u+1)dx$$ or tell me another approach.Thank you

Edit: My teacher has told me that this problem was not of three types I mentioned above, which was confusing me as I was set on it having to be one of them. Thank you everyone for your time and solutions.

Answer 1

In order to solve $$ xy'=yln(xy)$$

We make the substitution,

$$u = ln (xy)$$

Upon differentiation we get, $$du = (xdy + ydx)/(xy)$$ $$xy du = xdy + ydx $$

Divide by $ y$ and use the original differential equation, $ xy'=uy,$ to get: $$xdu=x/ydy+dx$$ $$xdy/dx=uy\rightarrow x/ydy=udx$$ $$ x du = (u + 1) dx$$

Solve the separable equation,
$$du / (u + 1) = dx / x$$ $$ln (u + 1) = ln x + const$$ $$u + 1 = kx$$ $$ln (xy) + 1 = kx $$ $$ln (xy) = kx - 1 $$ $$xy = exp(kx - 1)$$ $$y = exp(kx - 1) / x$$

Answer 2

Hint

$$xy'=y\ln(xy)$$ $$xy'+y=y\ln(xy)+y$$ $$xy'+y=y(\ln(xy)+1)$$ $$(xy)'=\frac {xy(\ln(xy)+1)} x$$ Integrate both sides $$\int \frac {d(xy)} {xy(\ln(xy)+1)}=\int \frac{dx}x$$ Substitute $z=xy$ for simplicity $$\int \frac {dz} {z(\ln(z)+1)}=\int \frac{dx}x$$ Substitute $z=e^t$ and ofcourse $dz=e^tdt$ $$\int \frac {\color{red}{e^t}dt} {\color{red}{e^t}(\ln(e^t)+1)}=\int \frac{dx}x$$ Note that $\ln(e^t)=t$ $$\int \frac {dt} {(t+1)}=\ln(x) +K$$ Don't forget to substitute back $t=\ln(z)=\ln(xy)$ $$\ln(t+1)=\ln(x)+K \implies t+1=Kx \implies \ln(xy)=Kx-1$$ And finally $$\boxed {y(x)=\frac {e^{Kx-1}} x}$$

Answer 3

Try the substitution $u = \ln(xy)$, so $u'=\frac{1}{x} +\frac{y'}{y}$, and after some algebra your equation becomes $u'x=u+1$ which is straightforward to solve.

Answer 4

That $\ln(xy)$ term really throws a monkey wrench into the works. If it were $\ln\left(\frac xy\right)$ the equation would be homogeneous and we would be home free. That suggests making the substitution $y=\frac1u$. Then $$x\frac{dy}{dx}=x\frac{d}{dx}\left(\frac1u\right)=-\frac x{u^2}\frac{du}{dx}=y\ln(xy)=\frac1u\ln\left(\frac xu\right)$$ Or $$\frac{du}{dx}=-\frac ux\ln\left(\frac xu\right)$$ Now this is homogeneous so we let $u=vx$: $$x\frac{dv}{dx}+v=-v\ln\left(\frac1v\right)=v\ln v$$ This is variables separable, so $$\frac{dv}{v\ln v-v}=\frac{dx}x$$ Now, we will be in a lot of trouble if the substitution $v=e^z$ doesn't work, so we try it and hold our breath: $$\frac{e^zdz}{e^z\cdot z-e^z}=\frac{dz}{z-1}=\frac{dx}x$$ Now we can integrate: $$\ln|z-1|=\ln|x|+C_1$$ $$\begin{align}z-1&=\pm e^{C_1}x=Cx=\ln v-1=\ln\left(\frac ux\right)-1=\ln u-\ln x-1\\ &=\ln \left(\frac1y\right)-\ln x-1=-\ln y-\ln x-\ln e\end{align}$$ Exponentiating, $$e^{Cx}=\frac1{exy}$$ $$y=\frac1xe^{-Cx-1}$$

Answer 5

Another approach to the given the ordinary differential equation

$xy' = y \ln (xy) \tag 1$

may be had by first dividing (1) through by $y$:

$x \dfrac{y'}{y} = \ln (xy), \tag 2$

then expanding out the right hand side

$x \dfrac{y'}{y} = \ln (xy) = \ln x + \ln y, \tag 3$

and recalling that

$\dfrac{y'}{y} = (\ln y)'; \tag 4$

thus (3) may be written

$x (\ln y)' = \ln y + \ln x; \tag 5$

inspection of (5) motivates the introduction of

$z = \ln y; \tag 6$

under this transformation of the dependent variable (5) becomes

$x z' = z + \ln x, \tag 7$

which after a little algebra takes the form

$z' - \dfrac{1}{x} z = \dfrac{\ln x}{x}. \tag 8$

We recognize (8) as a first order, linear, inhomogeneous ordinary differential equation of the type

$z' + P(x) z = Q(x); \tag 9$

the solution of (9) which takes the value $z(x_0) = \ln y(x_0)$ at $x = x_0$ is given by the well-known formula

$z(x) = \displaystyle \exp \left ( -\int_{x_0}^x P(s) \; ds \right ) \left ( z(x_0) + \int_{x_0}^x \exp \left ( \int_{x_0}^s P(t) \; dt \right ) Q(s) \; ds \right ); \tag{10}$

in the present case we see that

$P(x) = -\dfrac{1}{x}, \tag {11}$

and

$\displaystyle \int_{x_0}^x -\dfrac{1}{s} \; ds = - (\ln x -\ln x_0) = -\ln \dfrac{x}{x_0} = \ln \dfrac{x_0}{x}, \tag{12}$

whence

$\displaystyle \exp \left (-\int_{x_0}^x P(s) \; ds \right ) = \exp \left (-\int_{x_0}^x -\dfrac{1}{s} \; ds \right )$ $= \exp \left (-\ln \dfrac{x_0}{x} \right ) = \left(\exp \left (\ln \dfrac{x_0}{x} \right ) \right )^{-1} = \left (\dfrac{x_0}{x} \right )^{-1} = \dfrac{x}{x_0}; \tag{13}$

we also thus have

$Q(x) = \dfrac{\ln x}{x}, \tag{14}$

whence

$\displaystyle \int_{x_0}^x \exp \left ( \int_{x_0}^s P(t) \; dt \right ) Q(s) \; ds = \int_{x_0}^x \dfrac{x_0}{s} \dfrac{\ln s}{s} \; ds = x_0 \int_{x_0}^x \dfrac{\ln s}{s^2} \; ds; \tag{15}$

the right-most integral is susceptible to integration by parts; setting

$u = \ln s; \; dv = s^{-2}ds, \tag{16}$

we have

$du = s^{-1} \;ds; \; v = -s^{-1}; \tag{17}$

$\displaystyle \int_{x_0}^x \dfrac{\ln s}{s^2} \; ds = \int_{x_0}^x u \; dv = uv \mid_{x_0}^x - \int_{x_0}^x v \; du = -s^{-1} \ln s \mid_{x_0}^x - \int_{x_0}^x (-s^{-2}) \; ds$ $= \displaystyle \int_{x_0}^x (s^{-2}) \; ds - s^{-1} \ln s \mid_{x_0}^x = -s^{-1} \mid_{x_0}^x - s^{-1} \ln s \mid_{x_0}^x$$ = -\dfrac{1 + \ln s}{s} \mid_{x_0}^x = \dfrac{1 + \ln x_0}{x_0} - \dfrac{1 + \ln x}{x}; \tag{18}$

now returning to (15) we find

$\displaystyle \int_{x_0}^x \exp \left ( \int_{x_0}^s P(t) \; dt \right ) Q(s) \; ds = x_0 \left ( \dfrac{1 + \ln x_0}{x_0} - \dfrac{1 + \ln x}{x} \right ), \tag{19}$

and deploying (13) and (19) in (10) yields a solution for $z(x)$:

$z(x) = \dfrac{x}{x_0} \left (z(x_0) + x_0 \left ( \dfrac{1 + \ln x_0}{x_0} - \dfrac{1 + \ln x}{x} \right ) \right )$ $= \dfrac{z(x_0) x}{x_0} + x\left ( \dfrac{1 + \ln x_0}{x_0} - \dfrac{1 + \ln x}{x} \right ) = \dfrac{z(x_0) x}{x_0} + \dfrac{1 + \ln x_0}{x_0} x - (1 + \ln x)$ $ = \dfrac{z(x_0) + \ln x_0 + 1}{x_0} x - \ln x - 1; \tag{20}$

if we set

$K = \dfrac{z(x_0) + \ln x_0 + 1}{x_0}, \tag{21}$

then we may write

$z(x) = Kx - \ln x - 1; \tag{22}$

from (6) we have

$y(x) = e^{z(x)} = \exp(z(x)), \tag{23}$

whence

$y(x) = \exp(Kx - 1 - \ln x) = \exp(Kx - 1) \exp(-\ln x)$ $= \exp(Kx - 1) \exp(\ln x^{-1}) = \dfrac{\exp(Kx - 1)}{x}; \tag{24}$

we check:

$y'(x) = \dfrac{K \exp(Kx - 1) x - \exp(Kx - 1)}{x^2} = \dfrac{Kx - 1}{x^2} \exp(Kx - 1); \tag{25}$

$xy'(x) = \dfrac{Kx - 1}{x} \exp(Kx - 1); \tag{26}$

$xy(x) = \exp(Kx - 1); \tag{27}$

$y(x) \ln(xy(x)) = \dfrac{\exp(Kx - 1)}{x}(Kx - 1) = xy'(x). \tag{28}$

We have thus verified that (24) is the solution to (1).