1st-, 2nd-Order Diff-EQs: Find mu(y) for 2y^2(x+y^2)+xy(x+6y^2)(dy/dx=0, and hence...

[Wothom83]

4. Find an integrating factor \(\displaystyle \mu(y),\) depending only on y, for the differential equation:

. . . . .\(\displaystyle 2y^2\, (x\, +\, y^2)\, +\, xy\, (x\, +\, 6y^2)\, \dfrac{dy}{dx}\, =\, 0\)

...and hence find the general solution of this equation.

5. Find a polynomial \(\displaystyle p(x)\) so that \(\displaystyle p(x)\, e^{2x}\) is a solution of the second-order differential equation:

. . . . .\(\displaystyle \dfrac{d^2y}{dx^2}\, -\, 4\, \dfrac{dy}{dx}\, +\, 4y\, =\, x^2\, e^{2x}\)

...with the conditions that \(\displaystyle y(0)\, =\, 1\) and \(\displaystyle y(1)\, =\, 0.\)

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I need help with both questions. For the 4 I'm not sure how to start it, I've looked up how to do those questions but can't understand the formatting.
For 5 I set y=p(x)e^2x and differentiated to the point where I eliminated p(x) and p'(x) but am not sure where to go from there. Any help is appreciated

Thanks

 

[Deleted member 4993]

4. Find an integrating factor \(\displaystyle \mu(y),\) depending only on y, for the differential equation:

. . . . .\(\displaystyle 2y^2\, (x\, +\, y^2)\, +\, xy\, (x\, +\, 6y^2)\, \dfrac{dy}{dx}\, =\, 0\)

...and hence find the general solution of this equation.

5. Find a polynomial \(\displaystyle p(x)\) so that \(\displaystyle p(x)\, e^{2x}\) is a solution of the second-order differential equation:

. . . . .\(\displaystyle \dfrac{d^2y}{dx^2}\, -\, 4\, \dfrac{dy}{dx}\, +\, 4y\, =\, x^2\, e^{2x}\)

...with the conditions that \(\displaystyle y(0)\, =\, 1\) and \(\displaystyle y(1)\, =\, 0.\)

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I need help with both questions. For the 4 I'm not sure how to start it, I've looked up how to do those questions but can't understand the formatting.
For 5 I set y=p(x)e^2x and differentiated to the point where I eliminated p(x) and p'(x) but am not sure where to go from there. Any help is appreciated

Thanks

Can you please write the "general" form of the ODE that can be solved with integrating factor?

 

[HallsofIvy]

4. Find an integrating factor \(\displaystyle \mu(y),\) depending only on y, for the differential equation:

. . . . .\(\displaystyle 2y^2\, (x\, +\, y^2)\, +\, xy\, (x\, +\, 6y^2)\, \dfrac{dy}{dx}\, =\, 0\)

...and hence find the general solution of this equation.

5. Find a polynomial \(\displaystyle p(x)\) so that \(\displaystyle p(x)\, e^{2x}\) is a solution of the second-order differential equation:

. . . . .\(\displaystyle \dfrac{d^2y}{dx^2}\, -\, 4\, \dfrac{dy}{dx}\, +\, 4y\, =\, x^2\, e^{2x}\)

...with the conditions that \(\displaystyle y(0)\, =\, 1\) and \(\displaystyle y(1)\, =\, 0.\)

---

I need help with both questions. For the 4 I'm not sure how to start it, I've looked up how to do those questions but can't understand the formatting.
For 5 I set y=p(x)e^2x and differentiated to the point where I eliminated p(x) and p'(x) but am not sure where to go from there. Any help is appreciated

Thanks

An integrating factor for a differential equation of the form f(x,y)dx+ g(x,y)dy= 0 is a function u(x,y) such that f(x,y)u(x,y)dx+ g(x,y)u(x,y)dy= 0 is an "exact" equation. That is, we must have \(\displaystyle [f(x,y)u(x,y)]_y= [g(x,y)u(x,y)]_y\). Here, the differential equation is \(\displaystyle 3y^2(x+ y^2)dx+ xy(x+ 6y^2)dy= (3xy^2+ 3y^4)dx+ (x^2y+ 6xy^3)dy= 0\). \(\displaystyle (3xy^2+ 3y^4)_y= 6xy+ 12y^3\) and \(\displaystyle (x^2y+ 6xy^3)_x= 2xy+ 6y^3\). Since those involve only powers of x and y, I would try an integrating factor of the form \(\displaystyle x^my^n\). That is, we have \(\displaystyle (3x^{m+1}y^{n+2}+ 3x^my^{n+ 4})_y= 3(n+2)x^{m+1}y^{n+1}+ 3(n+4)x^my^{n+ 3}\) and \(\displaystyle (x^{m+2}y^{n+1}+ 6x^{m+1}y^{n+ 3})_x= (m+2)x^{m+1}y^{n+1}+ 6(m+1)x^my^{n+3}\). The powers of x and y are correct. In order that those be equal we must have 3(n+2)= m+2 and 3(n+4)= 6(m+ 1).

For problem 5, setting \(\displaystyle y= p(x)e^{2x}\) as suggested, \(\displaystyle y'= p'(x)e^{2x}+ 2p(x)e^{2x}\) and \(\displaystyle y''= p''(x)e^{2x}+ 4p'(x)e^{2x}+ 4p(x)e^{2x}\). Was that what you got? Putting those into the equation, \(\displaystyle p''(x)e^{2x}+ 4p'(x)e^{2x}+ 4p(x)e^{2x}- 4(p'(x)e^{2x}+ 2p(x)e^{2x})+ 4p(x)e^{2x}\)\(\displaystyle = (p''(x)+ 4p'(x)- 4p'(x)+ 4p(x)- 8p(x)+ 4p(x))e^{2x}\)\(\displaystyle = p''(x)e^{2x}= x^2e^{2x}\). In order for that to be true for all x, we must have \(\displaystyle p''(x)= x^2\). Integrate twice to determine p(x).