Y''*y' = x(x+1)

  • #1

Homework Statement


[itex]\frac{d^2 y}{dx^2}\cdot\frac{dy}{dx}=x(x+1), \hspace{10pt} y(0)=1, \hspace{5pt} y'(0)=2[/itex]


Homework Equations



None I can think of...

The Attempt at a Solution




The only thing I even thought to try was turn it into the form:
[itex]\frac{d^2 y}{dx^2}{dy}=x(x+1){dx}[/itex]
...and then it looks kind of like a separable equation?

I'm lost.
 

Answers and Replies

  • #3


That simplifies the equation to:
[itex]{w}\frac{dw}{dx}=x(x+1)[/itex]

Integrating both sides with respect to x gives:

[itex]\int{w}\frac{dw}{dx}dx=\int{x(x+1)}dx[/itex]
[itex]\frac{w^2}{2}=\frac{(\frac{dy}{dx})^2}{2}=\frac{x^3}{3}+\frac{x^2}{2}+C[/itex]

Now what? Am I on the right track?
 
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  • #4
798
1


[tex]wdw=(x^2+x)dx[/tex]
From here, you merely have to integrate both sides. You can utilize your initial conditions to determine a value for the constant of integration.


Edit: Yes, you are on the right track. Solve for C, then I am assuming you want to solve for y, however, you never really stated what is being asked.
 
  • #5


How can we use y(0)=1? Do we say:
[itex]\int_{x_o}^{0}wdx+C_2=1[/itex]?
 
  • #6
798
1


You don't need to use that condition. [itex][[y'(0)]^2/2=[x]^3/3+[x]^2/2+c_1[/itex] since [itex]y'=w[/itex]. Make sense? Can you take it from here?
 
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  • #7


You don't need to use that condition. [itex][[y'(0)]^2/2=[x(0)]^3/3+[x(0)]^2/2+c_1[/itex] since [itex]y'=w[/itex]. Make sense? Can you take it from here?

I don't understand why you said x(0)? I thought x was the independent variable? So why wouldn't it be:
[itex]\frac{y'(0)^2}{2}=\frac{2^2}{2}=2=\frac{0^3}{3}+ \frac{0^2}{2}+c_1[/itex]

... which means c1=2?

But I still don't get how that helps me find y?

that would just change the equation we want to solve to:

[itex]\frac{{[\frac{dy}{dx}]^2}}{2}=\frac{x^3}{3}+\frac{x^2}{2}+2[/itex]

But I still don't see how to find y from that. :(
 
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  • #8
798
1


You are right, that was a typo on my part. Get [itex]dy/dx[/itex] by itself. What is the relationship between y and y'?
 
  • #9


You are right, that was a typo on my part. Get [itex]dy/dx[/itex] by itself. What is the relationship between y and y'?

So multiply both sides by 2 and then take the square root? Getting:

[itex]\frac{dy}{dx}=\sqrt{\frac{2x^3}{3}+ x^2+4}[/itex]

And then integrate with respect to x...

I don't know how to do that integral, and wolframalpha couldn't do it either. (http://www.wolframalpha.com/input/?i=integrate+sqrt%282x^3%2F3%2Bx^2%2B4%29+with+respect+to+x)

Sorry to drag this out so long but I just don't get it!
 
  • #10
798
1


You don't have to necessarily evaluate the integral. You can change the integral to a definite integral; there will be an upper limit and a lower limit. You will change the variable of the integrand from x to some other variable. I can't integrate that either, so that is what I would do. If I missed something, I'm sure someone will jump in.

ie. [tex]\int{}\frac{e^{at^2}}{t^2}dt=\int_{s_0}^{s}\frac{e^{as^2}}{s^2}ds[/tex]
 
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  • #11


You don't have to necessarily evaluate the integral. You can change the integral to a definite integral; there will be an upper limit and a lower limit. You will change the variable of the integrand from x to some other variable. I can't integrate that either, so that is what I would do. If I missed something, I'm sure someone will jump in.

[itex]y(x)=\int_{x_o}^{x}\sqrt{\frac{2u^3}{3}+ u^2+4}du+C_2[/itex]
So
[itex]y(0)=\int_{x_o}^{0}\sqrt{\frac{2u^3}{3}+ u^2+4}du+C_2 = 1[/itex]
Choose xo=0, so
[itex]y(0)=\int_{0}^{0}\sqrt{\frac{2u^3}{3}+ u^2+4}du+C_2 = 0+C_2=1[/itex]
So C2=1, So the final answer is...

[itex]y(x)=\int_{0}^{x}\sqrt{\frac{2u^3}{3}+ u^2+4}du+1[/itex]



Is that right?
 

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