How can substitution be used to solve a differential equation?

In summary, the problem involves using substitution to solve for y in the equation y'=1/(x+y)^2-1. The solution involves substituting v=x+y and solving for y in terms of v, resulting in the answer y=[3(x-C)]^(1/3) - x. However, it's important to note that this is only one of three possible roots for the cubic equation, so this is the final answer unless given a specific point on the curve.
  • #1
scrtajntman
2
0

Homework Statement


The problem states: Use substitution to solve:

y'=1/(x+y)^2-1

Homework Equations


The Attempt at a Solution



I used the substitution of v=x+y

resulting in the answer y=[3(x-C)]^(1/3) - x

but I'm not too sure that's right

Can some help with the answer and the steps for getting it. It's just for a practice test so I'm not graded on it.
 
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  • #2
Show all of the steps you took to get the correct answer. If you made an error in any step, we could point it out then.
 
  • #3
scrtajntman said:

Homework Statement


The problem states: Use substitution to solve:

y'=1/(x+y)^2-1

Homework Equations


The Attempt at a Solution



I used the substitution of v=x+y

resulting in the answer y=[3(x-C)]^(1/3) - x

but I'm not too sure that's right

Can some help with the answer and the steps for getting it. It's just for a practice test so I'm not graded on it.

It looks like you got to [itex](x+y)^3=3x+C[/itex] correctly (you probably had -3C instead of +C but it makes no difference since both are just undetermined constants) ? ...If so, this is how you should leave your answer (unless you are given a point on the curve y(x)). The reason being is that there are actually three roots to this cubic equation, and [itex]y+x=\sqrt[3]{3x+C}[/itex] is only one of them.
 
  • #4
Great! So overall I got the problem right. Thanks.
 

What is integration by substitution?

Integration by substitution is a technique used to solve integrals by replacing the original variable with a new variable that simplifies the integrand. This method is especially useful for integrals involving trigonometric functions or expressions with complex exponents.

When should I use integration by substitution?

Integration by substitution is most commonly used when the integrand contains a function and its derivative, or when the integrand can be expressed in terms of a simpler function. This method is also helpful when the integrand contains a product or quotient of functions.

How do I choose the substitution variable?

The substitution variable is typically chosen to simplify the integrand. It should be a function that is easy to integrate and reduces the complexity of the expression. A good choice for the substitution variable is often the function that is inside a composite function, or a term in the integrand that is raised to a power.

What are the steps for integration by substitution?

The steps for integration by substitution are as follows:
1. Identify the substitution variable and its derivative in the integral.
2. Substitute the new variable and its derivative into the integral.
3. Simplify the integrand and rewrite the integral in terms of the new variable.
4. Integrate the new expression, keeping in mind to include the substitution variable in the final answer.
5. Substitute the original variable back into the solution to obtain the final answer.

Are there any common mistakes to avoid when using integration by substitution?

Yes, there are a few common mistakes to avoid when using integration by substitution. One is forgetting to include the derivative of the substitution variable in the integral. Another is using the wrong substitution variable, which can result in a more complicated integral. It is important to double check the substitution and its derivative to ensure the correct variable is being used. Lastly, forgetting to substitute the original variable back into the final answer can also result in an incorrect solution.

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