Solving a Differential Equation

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Homework Help Overview

The discussion revolves around solving a differential equation of the form x(dy/dx) - 4y = x^4e^x. Participants are attempting to find an appropriate method to solve this equation, exploring various approaches and interpretations of the problem.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the transformation of the original equation into a standard form and the calculation of the integrating factor. There are attempts to apply integration techniques and product rules, with some questioning the correctness of their steps and the resulting expressions.

Discussion Status

The discussion is ongoing, with participants sharing their attempts and questioning specific steps in their reasoning. Some guidance has been offered regarding the integrating factor, but there is no explicit consensus on the correctness of the approaches taken.

Contextual Notes

There are indications of confusion regarding the integration process and the nature of the resulting integrals, with references to advanced functions that may not align with the expectations of the participants' textbooks.

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Homework Statement



Solve the differential equation
x(dy/dx) -4y = x^4e^x

Homework Equations




The Attempt at a Solution




So this is what I did.

dy/dx -4y/x = x^3e^x

Then I did the integral of P(x) which I said was 4/x so the integral is lnx^4 then I(x) = e^lnx^4 = x^4. I then multiplied through by this and got


x^4(dy/dx) - 4x^3(y) = x^7e^x I said my product rule for my left side was (x^4(y))' so I ended up with x^4(y) = ∫ x^7(e^x) I just used a reduction formula for the integral. But is my procss OK? THanks
my equation b
 
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Jbreezy said:

Homework Statement



Solve the differential equation
x(dy/dx) -4y = x^4e^x

Homework Equations




The Attempt at a Solution




So this is what I did.

dy/dx -4y/x = x^3e^x

Then I did the integral of P(x) which I said was 4/x so the integral is lnx^4 then I(x) = e^lnx^4 = x^4. I then multiplied through by this and got

In this case [itex]P(x) = -4/x[/itex], so you've lost a minus sign in an exponent.

x^4(dy/dx) - 4x^3(y) = x^7e^x I said my product rule for my left side was (x^4(y))'
so I ended up with x^4(y) = ∫ x^7(e^x) I just used a reduction formula for the integral. But is my procss OK? THanks
my equation b

Try again with the correct [itex]P(x)[/itex].
 
Hey dude, So I did but it doesn't seem right I made P(x) as you say so my integrating factor become I(x) = 1/x^4

I multiply though and my expression becomes

(1/x^4)(dy/dx) - 4y/x^5 = e^x/x OK so the left side is then...

(y/x^4)' = e^x /x
But if you try and integrate the right side you have to do the err function. That can't be right because my book isn't that advanced. So Did I do something wrong?
 
Jbreezy said:
Hey dude, So I did but it doesn't seem right I made P(x) as you say so my integrating factor become I(x) = 1/x^4

I multiply though and my expression becomes

(1/x^4)(dy/dx) - 4y/x^5 = e^x/x OK so the left side is then...

(y/x^4)' = e^x /x
But if you try and integrate the right side you have to do the err function. That can't be right because my book isn't that advanced. So Did I do something wrong?

No, you did everything right. And yes, you don't get an elementary integral. It's not err it's Ei, but same problem.
 
OK. I kept looking over and it and I could not see where my math was wrong you know?
Hey I'm going to post another thread can you please help me ? It is solving a differential equation.
 
Jbreezy said:
OK. I kept looking over and it and I could not see where my math was wrong you know?
Hey I'm going to post another thread can you please help me ? It is solving a differential equation.

Sure, if I'm still around. Somebody else could probably handle it too.
 

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