Solution for these Differential equations

[Reshma]

Hi everyone. I'm trying to solve these differential equations but I could not crack a single nut. I seem to have lost my memory on solving differential equations . Please help me refresh it by providing useful hints.
I'm unable to separate the variables in the following. Perhaps I'm missing out on something important.

1] \frac{dy}{dx} + 2y = y^2e^{2x}

2] 2y\frac{dy}{dx} + y^2 = \frac{x}{2}e^{-x}

3] x^2\frac{dy}{dx} - 2xy = \frac{1}{x}

 

[saltydog]

Hi everyone. I'm trying to solve these differential equations but I could not crack a single nut. I seem to have lost my memory on solving differential equations . Please help me refresh it by providing useful hints.
I'm unable to separate the variables in the following. Perhaps I'm missing out on something important.

1] \frac{dy}{dx} + 2y = y^2e^{2x}

2] 2y\frac{dy}{dx} + y^2 = \frac{x}{2}e^{-x}

3] x^2\frac{dy}{dx} - 2xy = \frac{1}{x}

First one is a Riccati. Make the change of variables usually done for such equations and see what happens.

 

[Physics Monkey]

For the second one, can you simplfy the derivative term? After a well chosen substitution, the differential equation becomes linear.

For the third one, the equation is linear first order and there is a general method available.

 

[Reshma]

For the second one, can you simplfy the derivative term? After a well chosen substitution, the differential equation becomes linear.
For the third one, the equation is linear first order and there is a general method available.

Thank you so much for the help.

Well, I was able to solve the third one!

Bringing the equation in the general form:
\frac{dy}{dx} + P(x)y = Q(x)

\frac{dy}{dx} - \frac{2}{x} y = \frac{1}{x^3}

Setting y = u(x)v(x)

So,
\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}

u\left(\frac{dv}{dx} - \frac{2}{x}v\right) + v\frac{du}{dx} = \frac{1}{x^3}...(1)

Solving for v:
\frac{dv}{dx} - \frac{2}{x} v = 0

On solving:
\ln v = \ln x^2
v = x^2

Again on substituting for v in (1):
u = -\frac{1}{4x^4} + C

General formula:
y = v(x)\int \frac{Q(x)}{v(x)} dx + (C)v(x)

y = -\frac{1}{4x^2} + C{x}^2
Hope I'm right.
Sorry I could not find any suitable substitution for the second one . Please help!

 

[Reshma]

First one is a Riccati. Make the change of variables usually done for such equations and see what happens.

Ricatti? I haven't studied any differential equation like that. How do you solve such equations.

 

[saltydog]

Ricatti? I haven't studied any differential equation like that. How do you solve such equations.

First place it into standard form:

y^{'}+Q(x)y+R(x)y^2=P(x)

Now, make the transformation:

y=\frac{u^{'}}{Ru}

Can you now substitute this into the ODE? I'll start it for you:

y^{'}=\frac{Ruu^{''}-u^{'}(Ru^{'}+uR^{'})}{(Ru)^2}

right?

Make the other ones to get:

\frac{Ruu^{''}-u^{'}(Ru^{'}+uR^{'})}{(Ru)^2}+\frac{Qu^{'}}{Ru}+R\left(\frac{u^{'}}{Ru}\right)^2=P

Now simplify and obtain a second order in u. Solve, convert back to y, and I want a plot.

Edit: Suppose that last one looks a bit intimidating. That's just the general expression though. For your equation a lot of stuff just drops out leaving a simple second order to solve. Try it.

 

[Physics Monkey]

For the second one, focus on the term 2 y \frac{dy}{dx}, can you write this as something more convenient? Hint: notice that the only other y term you have is y^2.

 

[Reshma]

For the second one, focus on the term 2 y \frac{dy}{dx}, can you write this as something more convenient? Hint: notice that the only other y term you have is y^2.

I had tried it:

Set,
u = y^2

So,
\frac{du}{dx} = 2y\left(\frac{dy}{dx}\right)

So the eqaution becomes,
\frac{du}{dx} + u = \frac{x}{2} e^{-x}

But, I'm still unable to separate the variables. Should I adopt a different method?

 

[Benny]

I don't think you can separate variables. You could try the integrating factor technique since you have a first order linear differential equation.

 

[saltydog]

I had tried it:
Set,
u = y^2
So,
\frac{du}{dx} = 2y\left(\frac{dy}{dx}\right)
So the eqaution becomes,
\frac{du}{dx} + u = \frac{x}{2} e^{-x}
But, I'm still unable to separate the variables. Should I adopt a different method?

Reshma, rearrange the equation to:

2ydy+y^2dx=\frac{x}{2}e^{-x}dx

or:

\left(\frac{x}{2}e^{-x}-y^2\right)dx-2ydy=0

Now, we can make this exact right? You know, the partial of M with respect to y, partial of N with respect to x, do that arithmetic, get some function of x or y, then e to the integral of that function is the integrating factor right? You know this makes two plots now.

 

[Physics Monkey]

What Salty suggested is a nice way of proceeding, or as Benny said, you now have a linear first order equation in u and a general approach exists as I said before.

 

[Reshma]

Wow, thanks for your help, Saltydog and PhysicsMonkey. I got all the solutions!