Solving a Differential Equation

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In summary, the conversation is about solving a differential equation using an integrating factor. The person asks for confirmation on their process and the other person confirms that everything is correct, but the resulting integral is not elementary. They also offer to help with another differential equation problem.
  • #1
Jbreezy
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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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  • #2
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].
 
  • #3
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?
 
  • #4
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.
 
  • #5
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.
 
  • #6
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.
 

What is a linear differential equation?

A linear differential equation is an equation that involves a function and its derivatives, where the function and its derivatives are only raised to the first power. This means that the function and its derivatives are not multiplied together or raised to any higher powers.

What is the general form of a linear differential equation?

The general form of a linear differential equation is y' + p(x)y = g(x), where y' represents the first derivative of the function y, p(x) is a function of x, and g(x) is a function of x. This form can also be written as dy/dx + p(x)y = g(x).

What is the solution to a linear differential equation?

The solution to a linear differential equation is a function that satisfies the equation. In other words, when the solution function is substituted into the equation, it makes the equation true.

What is the order of a linear differential equation?

The order of a linear differential equation is the highest derivative of the function present in the equation. For example, the equation y'' + 2xy' + y = 0 would be a second order linear differential equation.

What are some applications of linear differential equations?

Linear differential equations are used in many fields of science and engineering, including physics, chemistry, biology, economics, and engineering. They can be used to model and predict various processes, such as population growth, radioactive decay, and heat transfer.

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