Solving a Diff Eq with Substitution

In summary: Applying the integrating factor should give you:y(t) = \frac{1}{\frac{1}{1+t^{2}}}[-\int\frac{2t}{(1+t^{2})(1+t^{2})}dt + c] y(t) = \frac{1}{\frac{1}{1+t^{2}}}[-\frac{2t}{1+t^{2}} + c] y(t) = 1+t^{2}[\frac{1}{1+t^{2}} + c] y(t) = 1 + c + ct^{2}So it looks like you just made a typo.
  • #1
KillerZ
116
0

Homework Statement



Solve the given differential equation by using appropriate substitution.

[tex]3(1+t^{2})\frac{dy}{dx} = 2ty(y^{3} - 1)[/tex]

Homework Equations



[tex]y = u^{\frac{1}{1-n}}[/tex]

The Attempt at a Solution



[tex]\frac{dy}{dt} = \frac{2ty^{4} - 2ty}{3 + 3t^{2}}[/tex]

[tex]\frac{dy}{dt} + \frac{2ty}{3 + 3t^{2}} = \frac{2ty^{4}}{3 + 3t^{2}}[/tex]

[tex]-\frac{1}{3}u^{-\frac{4}{3}}\frac{du}{dt} + (\frac{2ty}{3 + 3t^{2}})u^{-\frac{1}{3}} = (\frac{2ty^{4}}{3 + 3t^{2}})(u^{-\frac{1}{3}})^{4}[/tex]

[tex]-\frac{1}{3}\frac{du}{dt} + (\frac{2ty}{3 + 3t^{2}})\frac{u^{-\frac{1}{3}}}{u^{-\frac{4}{3}}} = \frac{2ty^{4}}{3 + 3t^{2}}[/tex]

[tex]-\frac{1}{3}\frac{du}{dt} + (\frac{2ty}{3 + 3t^{2}})u = \frac{2ty^{4}}{3 + 3t^{2}}[/tex]

[tex]\frac{du}{dt} - (\frac{2ty}{1 + t^{2}})u = -\frac{2ty^{4}}{1 + t^{2}}[/tex]

Integrating factor:

[tex]I(t) = e^{-\int\frac{2t}{1+t^{2}}dt}[/tex]

use substitution:

[tex]e^{ln|u|^{-1}}[/tex]

[tex]I(t) = \frac{1}{1+t^{2}}[/tex]

[tex]y(t) = \frac{1}{\frac{1}{1+t^{2}}}[-\int\frac{2t}{(1+t^{2})(1+t^{2})}dt + c][/tex]

[tex]y(t) = 1+t^{2}[\frac{1}{1+t^{2}} + c][/tex]

[tex]y(t) = 1 + c + ct^{2}[/tex]
 
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  • #2
For one thing, you shouldn't have t, u, and y all in the same equation. Here's an easier way to get started. Rewrite your equation:

3(1 + t2)y' + 2ty = 2ty4

Now multiply through by y-4:

3(1 + t2)y-4y' + 2ty-3 = 2t

Now let u = y-3, u' = -3y-4y'

This gets you to the equation

-(1 + t2)u' + 2tu = 2t

without all the fractional exponents and resulting errors.
 
  • #3
I made some typos in there here I fixed it:

Homework Statement



Solve the given differential equation by using appropriate substitution.

[tex]3(1+t^{2})\frac{dy}{dx} = 2ty(y^{3} - 1)[/tex]

Homework Equations



[tex]y = u^{\frac{1}{1-n}}[/tex]

The Attempt at a Solution



[tex]\frac{dy}{dt} = \frac{2ty^{4} - 2ty}{3 + 3t^{2}}[/tex]

[tex]\frac{dy}{dt} + \frac{2ty}{3 + 3t^{2}} = \frac{2ty^{4}}{3 + 3t^{2}}[/tex]

[tex]-\frac{1}{3}u^{-\frac{4}{3}}\frac{du}{dt} + (\frac{2t}{3 + 3t^{2}})u^{-\frac{1}{3}} = (\frac{2t}{3 + 3t^{2}})(u^{-\frac{1}{3}})^{4}[/tex]

[tex]-\frac{1}{3}\frac{du}{dt} + (\frac{2t}{3 + 3t^{2}})\frac{u^{-\frac{1}{3}}}{u^{-\frac{4}{3}}} = \frac{2t}{3 + 3t^{2}}[/tex]

[tex]-\frac{1}{3}\frac{du}{dt} + (\frac{2t}{3 + 3t^{2}})u = \frac{2t}{3 + 3t^{2}}[/tex]

[tex]\frac{du}{dt} - (\frac{2t}{1 + t^{2}})u = -\frac{2t}{1 + t^{2}}[/tex]

Integrating factor:

[tex]I(t) = e^{-\int\frac{2t}{1+t^{2}}dt}[/tex]

use substitution:

[tex]e^{ln|u|^{-1}}[/tex]

[tex]I(t) = \frac{1}{1+t^{2}}[/tex]

[tex]y(t) = \frac{1}{\frac{1}{1+t^{2}}}[-\int\frac{2t}{(1+t^{2})(1+t^{2})}dt + c][/tex]

[tex]y(t) = 1+t^{2}[\frac{1}{1+t^{2}} + c][/tex]

[tex]y(t) = 1 + c + ct^{2}[/tex]
 
  • #4
Interesting step that last one. Where did the denominator go?
 

1. What is substitution in differential equations?

Substitution is a technique used to simplify and solve a differential equation by replacing a variable with another expression or function. This allows us to transform the equation into a simpler form that can be solved using standard algebraic methods.

2. When should substitution be used to solve a differential equation?

Substitution is typically used when the differential equation is difficult to solve using traditional methods, such as separation of variables or integrating factors. It can also be used when the equation is in a specific form that lends itself well to substitution, such as the Bernoulli or exact form.

3. What steps should be followed when solving a differential equation with substitution?

The first step is to identify the form of the differential equation and determine if substitution is a viable method. Next, choose a suitable substitution, such as u = y/x or u = v(x), and substitute it into the equation. This will result in a new equation with only one variable. Finally, solve for the new variable and then substitute back in for the original variable to find the solution.

4. What are some common mistakes when using substitution to solve a differential equation?

One common mistake is choosing an inappropriate substitution, which can make the equation more complicated instead of simplifying it. Another mistake is not properly substituting back in for the original variable, which can lead to incorrect solutions. It is also important to check for extraneous solutions, as sometimes substitution can introduce them.

5. Are there any tips for using substitution effectively in solving differential equations?

It is helpful to practice identifying the form of differential equations and the appropriate substitution to use. It can also be useful to check the solution by differentiating it and plugging it back into the original equation. In some cases, it may be necessary to make multiple substitutions to fully solve the equation.

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