Integrating factor: x^2 (dy/dx) + 3x y = x^3 is multiplied by x...

[Vulcan]

Just starting on this subject but not following the attached!

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The equation

. . . . .\(\displaystyle x^2\, \dfrac{dy}{dx}\, +\, 3x\, y\, =\, x^3\)

is not exact. However, if we multiply it by x we obtain the equation

. . . . .\(\displaystyle x^3\, \dfrac{dy}{dx}\, +\, 3x^2\, y\, =\, x^4\)

This can be re-written as

. . . . .\(\displaystyle \dfrac{d}{dx}(x^3\,y)\, =\, x^4\)

which is an exact equation with solutuion

. . . . .\(\displaystyle x^3\, y\, =\, \)\(\displaystyle \displaystyle \int\, x^4\, dx\)

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Can someone explain how step three comes about from step 2
Also in step 4 where does the d/dx go? Should we not be integrating both sides?

As always, any help appreciated!

 

[Deleted member 4993]

Just starting on this subject but not following the attached!

---

The equation

. . . . .\(\displaystyle x^2\, \dfrac{dy}{dx}\, +\, 3x\, y\, =\, x^3\)

is not exact. However, if we multiply it by x we obtain the equation

. . . . .\(\displaystyle x^3\, \dfrac{dy}{dx}\, +\, 3x^2\, y\, =\, x^4\)

This can be re-written as

. . . . .\(\displaystyle \dfrac{d}{dx}(x^3\,y)\, =\, x^4\)

which is an exact equation with solutuion

. . . . .\(\displaystyle x^3\, y\, =\, \)\(\displaystyle \displaystyle \int\, x^4\, dx\)

---

Can someone explain how step three comes about from step 2
Also in step 4 where does the d/dx go? Should we not be integrating both sides?

As always, any help appreciated!

d/dx[x^3*y] = d/dx[x^3]*[y]+(x^3)*[dy/dx] .... applying product rule of differentiation

continue....

 

[Vulcan]

Sorry, still not getting it!

looks like you have integrated 3x^2?

 

[HallsofIvy]

No, Subhotosh Kahn did not integrate anything. Rather, he recognized that, using the product rule, \(\displaystyle \frac{d(x^3y)}{dx}= \frac{d(x^3)}{dx}y+ (x^3)\frac{dy}{dx}= 3x^2y+ x^3\frac{dy}{dx}\) so that you can go from \(\displaystyle 3x^2y+ x^3\frac{dy}{dx}\) to \(\displaystyle \frac{d(x^3y)}{dx}\).

 

[Vulcan]

I must be missing something!

 

[HallsofIvy]

Well, since you asked about a differential equation, I, at least, assumed you had taken Calculus! You do understand, don't you, that the derivative of \(\displaystyle x^3y\), with respect to x, is \(\displaystyle 3x^2y+ x^3\frac{dy}{dx}\). So you can replace \(\displaystyle x^3\frac{dy}{dx}+ 3x^2y\), in the equation, by \(\displaystyle \frac{d(x^3y}{dx}\).

 

[Vulcan]

Well, since you asked about a differential equation, I, at least, assumed you had taken Calculus! You do understand, don't you, that the derivative of \(\displaystyle x^3y\), with respect to x, is \(\displaystyle 3x^2y+ x^3\frac{dy}{dx}\). So you can replace \(\displaystyle x^3\frac{dy}{dx}+ 3x^2y\), in the equation, by \(\displaystyle \frac{d(x^3y}{dx}\).

Apologies, yes I have taken calculus. Just rushing through it too much and working on too many other things at the moment. Just looked back at my notes on implicit differentiation and have now got it. Thanks for your patience and assistance