Integration by separation of variables

In summary, the problem is to solve the differential equation dy/dt = 1/y^2 with the initial condition y(0) = 1 using variable separation. The solution is y = (3t+1)^1/3 and it can be found by moving the y^2 term to the left, integrating, and solving for y with the given initial condition.
  • #1
Physicist3
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0

Homework Statement



Using the technique involving variable separation, solve the following differential equation and use the initial condition to find the particular solution

[itex]\frac{dy}{dt}[/itex] = [itex]\frac{1}{y^{2}}[/itex] y(0) = 1

Homework Equations






The Attempt at a Solution



To be honest, could someone show me how to solve this one as I cannot seem to understand how to get to the book answer of y=(3t+1)[itex]^{1/3}[/itex] .

I understand that the y^2 should move to the left so that it becomes dy/y^2 but I am not sure how to progress with this?
 
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  • #2
Physicist3 said:

Homework Statement



Using the technique involving variable separation, solve the following differential equation and use the initial condition to find the particular solution

[itex]\frac{dy}{dt}[/itex] = [itex]\frac{1}{y^{2}}[/itex] y(0) = 1

Homework Equations






The Attempt at a Solution



To be honest, could someone show me how to solve this one as I cannot seem to understand how to get to the book answer of y=(3t+1)[itex]^{1/3}[/itex] .

I understand that the y^2 should move to the left so that it becomes dy/y^2 but I am not sure how to progress with this?
The y2 should be moved to the left side, but it doesn't become dy/y2.

What operation do you need to apply to both sides to get rid of the y2 in the denominator on the right?
 
  • #3
You should separate the variables, first move to the left the squared term, then move dt, now you must integrate don't forget the constant, now solve the equation for y. Y(0) it's a initial condition so when y = 0, t = 1, you should get a function with these conditions and this function ought to be the solution of the ODE.
 

1. What is Integration by Separation of Variables?

Integration by Separation of Variables is a method used in calculus to solve differential equations. It involves separating the variables in a differential equation and integrating each side with respect to one variable at a time.

2. When is Integration by Separation of Variables used?

Integration by Separation of Variables is used to solve differential equations that cannot be solved by other methods, such as substitution or integration by parts. It is commonly used in physics, engineering, and other fields that involve modeling and analyzing systems.

3. How does Integration by Separation of Variables work?

The method of Integration by Separation of Variables works by isolating the dependent and independent variables in a differential equation and then integrating each side separately. This allows for the differential equation to be transformed into an algebraic equation, which can be solved for the dependent variable.

4. What are the steps for Integration by Separation of Variables?

The steps for Integration by Separation of Variables are as follows:

  1. Identify the dependent and independent variables in the differential equation.
  2. Separate the variables on opposite sides of the equation.
  3. Integrate each side with respect to its respective variable.
  4. Apply any necessary initial conditions or boundary conditions.
  5. Solve the resulting algebraic equation for the dependent variable.

5. What are some limitations of Integration by Separation of Variables?

Integration by Separation of Variables can only be used to solve certain types of differential equations, specifically those that are separable. It may not work for more complex or nonlinear differential equations. Additionally, it may not always be possible to find an analytical solution using this method, which may require the use of numerical methods instead.

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