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Idealist10
 2 years ago
Find the general solution of (x2)(x1)y'(4x3)y=(x2)^3.
Idealist10
 2 years ago
Find the general solution of (x2)(x1)y'(4x3)y=(x2)^3.

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myininaya
 2 years ago
Best ResponseYou've already chosen the best response.1so i assumed you tried to put in the form y'+p*y=q form

Idealist10
 2 years ago
Best ResponseYou've already chosen the best response.1Please help me. How do I start? Divide (x2)?

Idealist10
 2 years ago
Best ResponseYou've already chosen the best response.1Okay, I'll try and tell me what's going on. Please don't leave.

myininaya
 2 years ago
Best ResponseYou've already chosen the best response.1you may have to some partial fractions just to let you know

Idealist10
 2 years ago
Best ResponseYou've already chosen the best response.1But how do I integrate (4x3)/((x2)(x1))?

Idealist10
 2 years ago
Best ResponseYou've already chosen the best response.1So do I multiply (x2)(x1)? It's x^23x+2.

myininaya
 2 years ago
Best ResponseYou've already chosen the best response.1\[\frac{34x}{(x2)(x1) }=\frac{A}{x2}+\frac{B}{x1}\] before you integrate you write what is on the left hand side in the right hand sides's form you must find A and B to complete that transaction

Idealist10
 2 years ago
Best ResponseYou've already chosen the best response.1So 4x/(x2)3/(x1)? And 4x(x1)3(x2)? But equal to what?

myininaya
 2 years ago
Best ResponseYou've already chosen the best response.1... Have you ever done partial fractions before?

Idealist10
 2 years ago
Best ResponseYou've already chosen the best response.1So isn't it 5ln(x2)ln(x1)?

myininaya
 2 years ago
Best ResponseYou've already chosen the best response.1I will show an example: \[\frac{1}{(x5)(x3)}\] pretend we wanted to write this as a sum of proper fractions we can attempt to do this by trying to write that fraction in this form A/(x5)+B/(x3) To find what A and B are we will need to combine those fractions in put in to form that is 1/[(x5)(x3)] form. so let's do that \[\frac{A}{x5}+\frac{B}{x3}=\frac{A(x3)+B(x5)}{(x5)(x3)}=\frac{(A+B)x3A5B}{(x5)(x3)}\] but the other side is equal to 1/[(x5)(x3)] so that means the other can have no x's so A+B would have to be 0 and the constant term will have to be equal to 1 therefore 3A5B=1 so we have a system of linear equations to solve: A+B=0 3A5B=1  To solve I will try to setup for elimination Multiply first equation by 3 3A+3B=0 3A5B=1 now add 2B=1 B=1/2 If B=1/2 and A+B=0, then A=1/2 So you know that \[\frac{1}{(x5)(x3)} \text{ can be written as } \frac{1}{2} \frac{1}{x5}\frac{1}{2}\frac{1}{x3}\] and we know how to intgrate both of those terms

myininaya
 2 years ago
Best ResponseYou've already chosen the best response.1@Idealist10 for the integrating factor there was a negative sign in front of y term so take what you have and multiply it by 1 but don't forget the integrate factor has that base e thing so for integrate factor we should have \[v=e^{5\lnx2+\lnx1}\] This can be a lot prettier

myininaya
 2 years ago
Best ResponseYou've already chosen the best response.1Now that was just one example above from partial fractions There are other things to consider when writing a fraction composed of polynomials as a sum of fractions

myininaya
 2 years ago
Best ResponseYou've already chosen the best response.1What I'm saying is the section on partial fractions can not be explained by one example

myininaya
 2 years ago
Best ResponseYou've already chosen the best response.1Anyways back to what you were saying...

myininaya
 2 years ago
Best ResponseYou've already chosen the best response.1to write your integrating factor a lot prettier you will need to recall some law of exponents and also some log properties

myininaya
 2 years ago
Best ResponseYou've already chosen the best response.1you do recall that x^(a+b) can be written as (x^a)(x^b) ?

myininaya
 2 years ago
Best ResponseYou've already chosen the best response.1do you see how that could be helpful here?

Idealist10
 2 years ago
Best ResponseYou've already chosen the best response.1So I got v=(x1)/(x2)^5, is this right?

Idealist10
 2 years ago
Best ResponseYou've already chosen the best response.1Let me work it out then.

Idealist10
 2 years ago
Best ResponseYou've already chosen the best response.1Yes, I got it! Thank you so much! Wow, this is a tough problem to solve!

myininaya
 2 years ago
Best ResponseYou've already chosen the best response.1Probably mostly because of the partial fractions part?

myininaya
 2 years ago
Best ResponseYou've already chosen the best response.1I guess you are in differential equations and try to recall all the things you learned in calculus 2. I do agree I had some trouble but with practice it should become easier on what you should do when you see something like that.

myininaya
 2 years ago
Best ResponseYou've already chosen the best response.1just practice practice it will sink in eventually
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