What is a particular solution to the differential equation $\frac{dy}{dx}=\frac{\mathrm{ln}x}{xy}$ and y(1)=2?

Corbin Bradford
2022-09-23
Answered

What is a particular solution to the differential equation $\frac{dy}{dx}=\frac{\mathrm{ln}x}{xy}$ and y(1)=2?

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asijikisi67

Answered 2022-09-24
Author has **10** answers

first separate it

so $y\frac{dy}{dx}=\frac{\mathrm{ln}x}{x}$

$\int dyy=\int dx\frac{\mathrm{ln}x}{x}$

$\frac{{y}^{2}}{2}=\frac{{\left(\mathrm{ln}x\right)}^{2}}{2}+C$

put in the IV

$\frac{{2}^{2}}{2}=\frac{{\left(\mathrm{ln}1\right)}^{2}}{2}+C\Rightarrow C=2$

$\frac{{y}^{2}}{2}=\frac{{\left(\mathrm{ln}x\right)}^{2}}{2}+2$

${y}^{2}={\left(\mathrm{ln}x\right)}^{2}+4$

so $y\frac{dy}{dx}=\frac{\mathrm{ln}x}{x}$

$\int dyy=\int dx\frac{\mathrm{ln}x}{x}$

$\frac{{y}^{2}}{2}=\frac{{\left(\mathrm{ln}x\right)}^{2}}{2}+C$

put in the IV

$\frac{{2}^{2}}{2}=\frac{{\left(\mathrm{ln}1\right)}^{2}}{2}+C\Rightarrow C=2$

$\frac{{y}^{2}}{2}=\frac{{\left(\mathrm{ln}x\right)}^{2}}{2}+2$

${y}^{2}={\left(\mathrm{ln}x\right)}^{2}+4$

asked 2022-09-29

What is a solution to the differential equation $\frac{dy}{dx}=(1+x)(1+y)$?

asked 2022-07-09

Find differential equation ${y}^{\mathrm{\prime}}=f(t,y)$ satisfied by $y(t)=4\phantom{\rule{thinmathspace}{0ex}}{e}^{2t}+3$

Solution:

Compute derivative of y,

${y}^{\mathrm{\prime}}=8\phantom{\rule{thinmathspace}{0ex}}{e}^{2t}$

Write right hand side above, in terms of the original function y, that is,

$y-3=4\phantom{\rule{thinmathspace}{0ex}}{e}^{2t}$

Get a differential equation satisfied by y, namely

${y}^{\mathrm{\prime}}=2y-6$

So my issue with that last answer. How is this a solution? Does it mean that if you somehow take an integral of $2y-6$ you should end up with the original $y(t)=4\phantom{\rule{thinmathspace}{0ex}}{e}^{2t}+3$???

It seems that there two different derivatives of y(t)

one is:

${y}^{\mathrm{\prime}}=8\phantom{\rule{thinmathspace}{0ex}}{e}^{2t}$

the other is:

${y}^{\mathrm{\prime}}=2y-6$

and I don't get it, can someone explain?

Also a bit offtopic, but the way y'=f(t,y) is written kinda bugs me.

Shouldn't it be written like y'=f(t,y(t)) to show that the function f contains t as an independent variable and the function y(t) which contains variable t as an input to itself (dependent variable t) ??? That's kinda an essential information, so surprised it's omitted in the writings.

Solution:

Compute derivative of y,

${y}^{\mathrm{\prime}}=8\phantom{\rule{thinmathspace}{0ex}}{e}^{2t}$

Write right hand side above, in terms of the original function y, that is,

$y-3=4\phantom{\rule{thinmathspace}{0ex}}{e}^{2t}$

Get a differential equation satisfied by y, namely

${y}^{\mathrm{\prime}}=2y-6$

So my issue with that last answer. How is this a solution? Does it mean that if you somehow take an integral of $2y-6$ you should end up with the original $y(t)=4\phantom{\rule{thinmathspace}{0ex}}{e}^{2t}+3$???

It seems that there two different derivatives of y(t)

one is:

${y}^{\mathrm{\prime}}=8\phantom{\rule{thinmathspace}{0ex}}{e}^{2t}$

the other is:

${y}^{\mathrm{\prime}}=2y-6$

and I don't get it, can someone explain?

Also a bit offtopic, but the way y'=f(t,y) is written kinda bugs me.

Shouldn't it be written like y'=f(t,y(t)) to show that the function f contains t as an independent variable and the function y(t) which contains variable t as an input to itself (dependent variable t) ??? That's kinda an essential information, so surprised it's omitted in the writings.

asked 2022-09-13

How do you find all solutions of the differential equation $\frac{{d}^{2}y}{{dx}^{2}}={x}^{-2}$?

asked 2022-06-24

Given the first-order differential equation:

$\frac{dy}{dx}=-6xy$

The textbook says you cannot differentiate both sides as y is on the right side and you have to use separation of variables. However, I did integrate both sides and arrived at this:

$\int \frac{dy}{dx}dx=\int -6xydx$

$y=-6y\int xdx$

$y=-6y(\frac{{x}^{2}}{2}+C)$

$y=-3y{x}^{2}+C$

$y+3y{x}^{2}=C$

$y(1+3{x}^{2})=C$

$y(x)=\frac{C}{1+{x}^{2}}$

And the second last step is valid since $1+3{x}^{2}$ can never be zero. However, this is not the correct answer, which is:

$y(x)=C{e}^{-3{x}^{2}}$. Why is this?

$\frac{dy}{dx}=-6xy$

The textbook says you cannot differentiate both sides as y is on the right side and you have to use separation of variables. However, I did integrate both sides and arrived at this:

$\int \frac{dy}{dx}dx=\int -6xydx$

$y=-6y\int xdx$

$y=-6y(\frac{{x}^{2}}{2}+C)$

$y=-3y{x}^{2}+C$

$y+3y{x}^{2}=C$

$y(1+3{x}^{2})=C$

$y(x)=\frac{C}{1+{x}^{2}}$

And the second last step is valid since $1+3{x}^{2}$ can never be zero. However, this is not the correct answer, which is:

$y(x)=C{e}^{-3{x}^{2}}$. Why is this?

asked 2021-01-28

Solve differential equation ${y}^{\prime}+2xy=4x$

asked 2022-07-01

I was solving for the equation of a curve and I arrived at the following differential equation. However I do not know how to solve it:

$\frac{dy}{dx}=\frac{-x+\sqrt{{x}^{2}+{y}^{2}}}{y}$

$\frac{dy}{dx}=\frac{-x+\sqrt{{x}^{2}+{y}^{2}}}{y}$

asked 2022-07-01

Consider the linear first order non-homogeneous partial differential equation

${U}_{x}+y{U}_{y}-y=y{e}^{-x}$

By using the method of characteristics show that its general solution is given by

$u(x,y)=-y{e}^{-x}+{e}^{x}g(y{e}^{-x})$

where g is any differentiable function of its argument

$A=1,B=y,d=-1,f=y{e}^{-x}\phantom{\rule{0ex}{0ex}}{\displaystyle \frac{dy}{dx}}={\displaystyle \frac{b}{a}}=y\phantom{\rule{0ex}{0ex}}dy=ydx\phantom{\rule{0ex}{0ex}}\int {\displaystyle \frac{1}{y}}dy=\int dx\phantom{\rule{0ex}{0ex}}\mathrm{ln}y=x+c\phantom{\rule{0ex}{0ex}}\eta (x,y)=c\phantom{\rule{0ex}{0ex}}\text{let}\xi =x$

Then

$\eta =\mathrm{ln}y-x$

$\xi =x$

those are my characteristics curves,

when I put it into $[a{\xi}_{x}+b{\xi}_{y}]{\omega}_{\xi}+d\omega =F$

I get the equation ${\omega}_{\xi}-\omega ={e}^{\eta}$

Is this all correct?

${U}_{x}+y{U}_{y}-y=y{e}^{-x}$

By using the method of characteristics show that its general solution is given by

$u(x,y)=-y{e}^{-x}+{e}^{x}g(y{e}^{-x})$

where g is any differentiable function of its argument

$A=1,B=y,d=-1,f=y{e}^{-x}\phantom{\rule{0ex}{0ex}}{\displaystyle \frac{dy}{dx}}={\displaystyle \frac{b}{a}}=y\phantom{\rule{0ex}{0ex}}dy=ydx\phantom{\rule{0ex}{0ex}}\int {\displaystyle \frac{1}{y}}dy=\int dx\phantom{\rule{0ex}{0ex}}\mathrm{ln}y=x+c\phantom{\rule{0ex}{0ex}}\eta (x,y)=c\phantom{\rule{0ex}{0ex}}\text{let}\xi =x$

Then

$\eta =\mathrm{ln}y-x$

$\xi =x$

those are my characteristics curves,

when I put it into $[a{\xi}_{x}+b{\xi}_{y}]{\omega}_{\xi}+d\omega =F$

I get the equation ${\omega}_{\xi}-\omega ={e}^{\eta}$

Is this all correct?