Is this the correct answer for a partial differentiation question?

In summary, the conversation discusses finding the general solution of a differential equation, specifically y' + (2/x)y = 3/(x^2). The person attempting the solution makes a mistake in their initial approach but eventually arrives at the correct solution by using the integrating factor method. They also mention their struggles with studying mathematics to become a high school math teacher.
  • #1
laura_a
66
0

Homework Statement



Find the general solution of

y' + (2/x)y = 3/(x^2)



The Attempt at a Solution




xy' + 2y = 3/x
d/dx (x * 2y) = 3/x
integrating both sides (using product rule for LHS) I end up with

y= (3lnx + C)/2x

Then I am supposed to find the solution for which y(2) = -1 so I subbed that into the above and got c = -4/3 - ln2

So then I can say that y = 3lnx/(2x) - 4/3 -ln2

Hope that's right!
Thanks guys!
 
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  • #2
your first step was right
u shud have repeated that
(x^2)y' + 2xy = 3
d/dx ((x ^ 2)*y) = 3
integrating both sides

y(x^2)+c=3x
this is the general solution now u can find c by given values of x and y and not supposing.
 
  • #3
laura_a said:

Homework Statement



Find the general solution of

y' + (2/x)y = 3/(x^2)
In what sense is this a partial differential equation?



The Attempt at a Solution




xy' + 2y = 3/x
d/dx (x * 2y) = 3/x
These are NOT the same. differentiating (x*2y) gives 2xy'+ 2y. there is no "2" multipying xy' in the equation.

integrating both sides (using product rule for LHS) I end up with
There is no "product rule" for integration. You are simply using the "fundamental theorem of calculus, that the integral of f ' is f.

y= (3lnx + C)/2x
If y= (3lnx+ C)/2x, then y'= [2x(3/x)- 2(3lnx+C)]/4x2
Then xy'+ 2y= [6- 6lnx+2C]/4x+ [6lnx+ 2C/4x= 3/2x, not 3/x.

Then I am supposed to find the solution for which y(2) = -1 so I subbed that into the above and got c = -4/3 - ln2

So then I can say that y = 3lnx/(2x) - 4/3 -ln2

Hope that's right!
Thanks guys!
As I pointed out d/dx[2xy] is NOT equal to xy'+ 2y. There is NO function of x and y whose derivative with respect to x is xy'+ 2y, it is not an exact derivative. However, multiplying the entire equation by the "integrating factor", x, wil make it exact.

Multiplying the equation xy'+ 2y= 3/x by x gives x2y'+ 2xy= 3. The left side is equal to d[x2y]/dx so the equation is the same as d(x2y)/dx= 3 and integrating both sides of that x2y= 3x+ C.
 
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  • #4
THanks guys, I know I typed partial but it is not what I meant... I've read about 16 textbooks today I think and I've made myself more confused... I understand the part that I missed now and I didn't even think to differentiate and check if my answer was right, I was more surprised that I got an answer... I'm studying very hard mathematics in order to teach high school maths and I'm struggling... thanks heaps for all your info
 

1. What is partial differentiation?

Partial differentiation is a mathematical concept used in multivariable calculus to describe the rate of change of a function with respect to one of its variables while holding all other variables constant.

2. How do you find the partial derivative of a function?

To find the partial derivative of a function, you take the derivative with respect to one variable while treating all other variables as constants. This can be done using standard differentiation rules.

3. What is the purpose of partial differentiation?

Partial differentiation is used to analyze the behavior of a function in multiple dimensions. It can help us understand how a function changes along different directions and identify critical points such as maxima and minima.

4. Are there any special cases for partial differentiation?

Yes, there are special cases such as when a function is not continuous or differentiable at a certain point, or when the function involves transcendental functions. In these cases, special techniques like the chain rule may be necessary to find the partial derivative.

5. How is partial differentiation different from ordinary differentiation?

Ordinary differentiation involves finding the derivative of a function with respect to one variable. Partial differentiation, on the other hand, involves finding the derivative of a function with respect to one variable while treating all other variables as constants. This allows us to analyze the behavior of a function in multiple dimensions.

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