Solving this Linear Differential Equation?

In summary: Then du=dx and v=\frac{1}{18}e^{x^{18}}\int e^{x^{18}}x^{18}\,dx=x^{18}\int e^{x^{18}}\,dx-\int x^{17}e^{x^{18}}\,dx=x^{18}e^{x^{18}}-\frac{1}{18}e^{x^{18}}=e^{x^{18}}(x^{18}-\frac{1}{18})+CTherefore, the general solution is y=\frac{e^{x^{18}}(x^{18}-\frac{1}{18})+C}{e^{x^{18}}}=\frac{x^{18
  • #1
τheory
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Homework Statement


Find the general solution to this differential equation:
[itex]\frac{dy}{dx}+18x^{17}y=x^{18}[/itex]
Use the variable [itex]I[/itex] for replacing [itex]\int e^{x^{18}} dx[/itex]

2. The attempt at a solution
I've solved the differential equation using integrating factors and obtained the following result:
[itex]y=\frac{\int e^{x^{18}}x^{18} dx + C}{e^{x^{18}}}[/itex]

My problem is using the variable [itex]I[/itex] for replacing the [itex]\int e^{x^{{18}}} dx[/itex]. How am I suppose to do this with [itex]x^{18}[/itex] inside the integral?
 
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  • #2
You need to group characters using { } in LaTeX.
τheory said:

Homework Statement


Find the general solution to this differential equation:
[tex]\frac{dy}{dx}+18x^{17}y=x^{18}[/tex]Use the variable [itex]I[/itex] for replacing [itex]\int e^{x^{18}}\,dx[/itex]

2. The attempt at a solution
I've solved the differential equation using integrating factors and obtained the following result:
[tex]y=\frac{\int e^{x^{18}}x^{18}\,dx + C}{e^{x^{18}}}[/tex]My problem is using the variable [itex]I[/itex] for replacing the [itex]\int e^{x^{18}}\,dx[/itex]. How am I suppose to do this with x^18 inside the integral?
You can't. You want to integrate by parts now.
 
  • #3
Are you familiar with the product rule?

(f*g)' = f'g + g'f

I am not following what you are doing at all...

You already got the integrating factor [tex]e^{x^18}[/tex], why do you have [tex]\int e^{x^18} dx[/tex]?
 
  • #4
Okay well I did try using integration by parts after I obtained the general solution, but didn't get far as I got this:
[itex]\int e^{x^{18}}x^{18}dx[/itex]

[itex]u = x^{18}[/itex]
[itex]du = 18x^{17} dx[/itex]
[itex]dv = e^{x^{18}} dv[/itex]
[itex]v = \int e^{x^{18}} dv[/itex]

[itex]uv - \int v du[/itex]

[itex]x^{18}\int e^{x^{18}} dv - \int[ \int e^{x^{18}} dv] 18x^{17} dx[/itex]

At this point, how am I suppose to utilize the substitution of [itex]\int e^{x^{18}}dx[/itex] with [itex]I[/itex]? In other words, how do I input the general solution into the website without inputting integrals? Since the website doesn't recognize integral signs, which is why it's asking me to use [itex]I[/itex] for every [itex]\int e^{x^{18}}dx[/itex] I encounter.

To the flyingpig's question, I need to use the variable "I" because it says to do so in the online problem that I'm doing.
 
  • #5
Try using u=x and [itex]dv=x^{17}e^{x^{18}}\,dx[/itex].
 

FAQ: Solving this Linear Differential Equation?

What is a linear differential equation?

A linear differential equation is an equation that involves a function and its derivatives, with the highest derivative having a power of 1. It can be written in the form of y' + p(x)y = q(x), where y is the dependent variable and x is the independent variable.

How do you solve a linear differential equation?

To solve a linear differential equation, you need to follow a series of steps. First, rearrange the equation into the form of y' + p(x)y = q(x). Then, find the integrating factor, which is e∫p(x)dx. Next, multiply the integrating factor to both sides of the equation and integrate both sides. Finally, solve for y(x) by using the initial conditions.

What are the initial conditions in a linear differential equation?

The initial conditions in a linear differential equation are the values of the dependent variable and its derivatives at a specific point, usually denoted as x0. These initial conditions are necessary to find the particular solution of the differential equation.

Can a linear differential equation have more than one solution?

Yes, a linear differential equation can have more than one solution. This is because when integrating both sides of the equation, an arbitrary constant C is introduced. The value of this constant can change the solution of the equation.

What is the difference between a linear and a non-linear differential equation?

A linear differential equation has the dependent variable and its derivatives raised to the first power, while a non-linear differential equation can have higher powers. This makes linear differential equations easier to solve compared to non-linear ones. Additionally, the solutions of linear differential equations can be added together to form a general solution, while this is not possible for non-linear equations.

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