Separable differential equation

In summary: That means (-9) (-9) = 81. That is the same as + 81. In summary, the given differential equation is separable, and after solving for y, the initial condition is used to find the value of the constant. However, when plugging in the initial condition, it is important to use the correct variable and to be careful with the signs when squaring a negative number.
  • #1
TheKracken
356
7

Homework Statement


2y * y'(t) = 3t^2 such that y(0) = 9

Homework Equations


g(y)y'(t) = h(t)

The Attempt at a Solution



So I have done many of these seperable ones in homework that did not require a parameter so now I got lost.
This is what I did;[/B]
Integral of 2y dy = integral of 3t^2 dt

y^2 + C = t^3

Then since y(0) = 9 I assumed I plug in 0 for y and that gives me 0 + C= t^3
so I figured that means that C= 9

then solving for y I get

y= sqrt (t^3 - 9) but that is not the answer. It is sqrt(t^3 +81) now that sort of is similar because -9^2 is 81 but I don't see any correlation to any mistakes. Any help?
 
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  • #2
It looks like you did everything right, until plugging in zero. zero is ##t=0## not ##y=0##.
 
  • #3
TheKracken said:

Homework Statement


2y * y'(t) = 3t^2 such that y(0) = 9

Homework Equations


g(y)y'(t) = h(t)

The Attempt at a Solution



So I have done many of these seperable ones in homework that did not require a parameter so now I got lost.
This is what I did;[/B]
Integral of 2y dy = integral of 3t^2 dt

y^2 + C = t^3
I would put the constant on the right side, but it should work this way, too.
TheKracken said:
Then since y(0) = 9 I assumed I plug in 0 for y and that gives me 0 + C= t^3
No. The initial condition means that y = 9 when t = 0.
TheKracken said:
so I figured that means that C= 9

then solving for y I get

y= sqrt (t^3 - 9) but that is not the answer. It is sqrt(t^3 +81) now that sort of is similar because -9^2 is 81 but I don't see any correlation to any mistakes. Any help?
-9^2 means the negative of 92, or - 81. To square -9, write it as (-9)2, with parentheses.
 

1. What is a separable differential equation?

A separable differential equation is a type of differential equation in which the variables can be separated on either side of the equals sign. This means that the dependent variable and its derivative are on opposite sides of the equation.

2. How do you solve a separable differential equation?

To solve a separable differential equation, you need to separate the variables on either side of the equals sign, integrate both sides, and then solve for the constant of integration. This will give you the general solution, which can then be used to find the particular solution by plugging in the given initial conditions.

3. What are the advantages of using separable differential equations?

One advantage of using separable differential equations is that they can be solved using basic integration techniques, making them more accessible for beginners in calculus. Separable differential equations also have many practical applications in fields such as physics, engineering, and economics.

4. Can all differential equations be solved using separation of variables?

No, not all differential equations can be solved using separation of variables. This method only works for certain types of differential equations, specifically those that are linear and can be separated into two distinct functions.

5. How are separable differential equations used in real-life situations?

Separable differential equations are commonly used to model the behavior of natural phenomena, such as population growth, radioactive decay, and chemical reactions. They can also be used to solve problems related to motion, heat transfer, and electrical circuits.

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