Problem of differential equation

[esmeco]

Homework Statement

I'm having a problem solving this equation!I'm stuck at a part of the equation where I don't know wha to do!This's the problem:

Show that the given function satisfy the indicated differential equation:

Homework Equations

y=cx +c -c^2; (dy/dx)^2 - dy/dx -x(dy/dx) + y=0

The Attempt at a Solution

The attempt to the solution is on the image below

 

[cristo]

Just plug the function into the differential equation. If c and c² are constants, then what can you say about their derivatives?

I've not looked at your figure as its pending approval, so apologies if that's what you've tried. If so, check your arithmetic, as a mental calculation suggests that it works.

 

[HallsofIvy]

I did look at your attachement and it's hard to see what you are trying to do! In particular, there is a "u" in your equations that does not appear in the problem.
If y= cx+ c²- c, then y'= c and y"= 0.
The difficulty may be that y= cx+ c²- c does NOT satisfy
y"- y'- xy'+ y= 0!

What, exactly, does the problem say?

 

[cristo]

I did look at your attachement and it's hard to see what you are trying to do! In particular, there is a "u" in your equations that does not appear in the problem.
If y= cx+ c²- c, then y'= c and y"= 0.
The difficulty may be that y= cx+ c²- c does NOT satisfy
y"- y'- xy'+ y= 0!

What, exactly, does the problem say?

The OP writes the equation as (y')²- y'- xy'+ y= 0. This equation is satisfied by y= cx+ c²- c. Perhaps this is causing confusion. I have not seen the work, however, so am just wildly guessing where the problem may lie!

 

[esmeco]

Thank you for all the replies!Oh by the way, HallsofIvy, you mistyped the equation it should be y=cx +c -c^2 and not y= cx+ c^2- c like you have (although this can easily be mistyped)...Well, regarding the problem,I'm trying to prove that, by solving this equation: (dy/dx)^2 - dy/dx -x(dy/dx) + y=0 , we get a result like this y= cx+ c^2- c .Like Cristo said we get an equation
like that (y')^2- y'- xy'+ y= 0 (which also appears in my resolution), but I'm not sure how should I go from there...Like we have different types of differential equations and different methods of solving those types of equations,so, what type of equation (linear 1st order,bernoulli,separate variables,etc.) does this equation belong to?I'm out of ideas right now...Thanks in advance for the reply!

 

[cristo]

Thank you for all the replies!Oh by the way, HallsofIvy, you mistyped the equation it should be y=cx +c -c^2 and not y= cx+ c^2- c like you have (although this can easily be mistyped)...Well, regarding the problem,I'm trying to prove that, by solving this equation: (dy/dx)^2 - dy/dx -x(dy/dx) + y=0 , we get a result like this y= cx+ c^2- c .Like Cristo said we get an equation
like that (y')^2- y'- xy'+ y= 0 (which also appears in my resolution), but I'm not sure how should I go from there...Like we have different types of differential equations and different methods of solving those types of equations,so, what type of equation (linear 1st order,bernoulli,separate variables,etc.) does this equation belong to?I'm out of ideas right now...Thanks in advance for the reply!

Your original question says: Show that y=.. satisfies the given differential equation. So, the simple way to do this, is to plug it into the equation. I.e. find y' and square it for the first term.. etc. You should then have an equation redundant of terms in y. If the LHS of this equation equals 0(the RHS), then you have shown that y=.. satisfies the differential equation.

Try this. If you can't get it post your attempts and I'll try and find your mistake.

 

[esmeco]

Now I get it...So the equation would be like: (y')^2 - y' -x*y' +c*x+c-c^2=0

and y'=c , so making the substitution we have: c^2 - c - X*c +c*x +c -c^2=0 and 0=0 true.
I got confused with this simple equation because I have demonstrated another equation (with the same statement) where I had to demonstrate that y=senx-1 +ce^-senx satisfied this equation : dy+ycosx=sen(2x)/2 .I have demonstrated this in another way by using this equation: y' + p(x)*y=q(x) .
Is there another way to demonstrate the equation I solved besides substitution(for example solving it like I solved the one I just talked about)?Thanks to all for the help!

 

[cristo]

Now I get it...So the equation would be like: (y')^2 - y' -x*y' +c*x+c-c^2=0

and y'=c , so making the substitution we have: c^2 - c - X*c +c*x +c -c^2=0 and 0=0 true.

Correct

I got confused with this simple equation because I have demonstrated another equation (with the same statement) where I had to demonstrate that y=senx-1 +ce^-senx satisfied this equation : dy+ycosx=sen(2x)/2 .I have demonstrated this in another way by using this equation: y' + p(x)*y=q(x) .

The equation y'+ycosx=sin(2x)/2 is a first order linear differential equation. We can solve such equations by expressing them in the form y' + p(x)*y=q(x) and using the integrating factor method (which I think is what you mean).

Is there another way to demonstrate the equation I solved besides substitution(for example solving it like I solved the one I just talked about)?Thanks to all for the help!

The differential equation you give in your original question cannot be expressed in the form y' + p(x)*y=q(x), and thus the integating factor method cannot be used. I can't think of a way, off the top of my head, of how to solve the original differential equation, apart from using trial and error, starting with the trial function y=cx, and working up from that. However, since the question did not ask you to solve the equation; verifying the solution given is sufficient.

 

[esmeco]

I guess that was my problem with the equation...I was tring to solve it like a first order linear differential equation which is not possible...Thanks for the help!:D

 

[HallsofIvy]

If you were asked to show that x= 2 is a solution to the equation x⁵- 3x⁴+ x²+ 10x+ 2= 0 would you try to solve the equation? It is much simpler to just replace x with 2 and show that the equation is true for x= 2. It is always much easier to show that a given function satisfies a differential equation that it is to solve the equation!