Differential Equations

 

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Separable Differential Equations

In this section we will evaluate separable differential equations. Looking more closely at that term: we will be looking at equations that have \frac{dy}{dx} in them, and finding the general antiderivative. The separable part of the term means that the x and y variables can easily be separated from each other.
Steps for solving Separable Differential Equations:
1. Separate the equation so that all occurrences of y are on the left side of the =, and all occurrences of x are on the right side.
2. Take the antiderivative of both sides.
3. Solve for y
Example 1:
Consider the equation:

\frac{dy}{dx} = \frac{x^2}{y}

In order to separate the x and y from each other, you will multiply both sides by y and by dx:

y dy = x^2 dx

Integrating both sides:  (This is legitimate, as long as it is done to both sides of the equation.)

\int y dy = \int x^2 dx

\frac{1}{2} y^2 = \frac{1}{3} x^3 + C

(You may think it is necessary to add a constant to both sides: \frac{1}{2} y^2 + C_1= \frac{1}{3} x^3 + C_2, but then you can combine the constants y^2 = \frac{1}{3} x^3 + C_2 - C_1, and call the new constant C_3, or just C. This will make things easier and more tidy.)
So we have:
\frac{1}{2} y^2 = \frac{1}{3} x^3 + C, and want to solve for y. Multiplying both sides by 2:

y^2 = \frac{2}{3} x^3 + 2C, or y^2 = \frac{2}{3} x^3 + C

(Note that 2C is also just a constant, so you can just call that C as well.)
Taking the square root of both sides:

y = \sqrt{\frac{2}{3} x^3 + C}

Checking your answer:
You may be in the habit (a good habit!) of checking your integral by taking the derivative of it, in order to make sure that it matches the original problem. In this case, the derivative will look nothing like the original problem – but it actually is the same!
Taking the derivative of your answer:

\frac{dy}{dx} = \frac{d}{dx} \left( \sqrt{\frac{2}{3} x^3 + C} \right)= \frac{d}{dx}\left(\frac{2}{3} x^3 + C \right)^{\frac{1}{2}}

(Recall that roots can be re-written as rational exponents, and this makes them easier to integrate and differentiate.)

\frac{dy}{dx} = \frac{1}{2}\left(\frac{2}{3} x^3 + C\right)^{\frac{-1}{2}}(2 x^2) = x^2 \left(\frac{2}{3} x^3 + C \right)^{\frac{-1}{2}}

(Remember to use the chain rule!)
Th original problem was: \frac{dy}{dx} = \frac{x^2}{y}. Can these really be the same thing? Set them equal to each other and see:

\frac{x^2}{y} \overset{?}{=} x^2 \left(\frac{2}{3} x^3 + C \right)^{\frac{-1}{2}}

Replacing y on the left side with y = \sqrt{\frac{2}{3} x^3 + C} from above, we get:

\frac{x^2}{\sqrt{\frac{2}{3} x^3 + C}} \overset{\checkmark}{=} x^2 \left(\frac{2}{3} x^3 + C \right)^{\frac{-1}{2}}

Since the fractional exponent gives the square root, and the negative in the exponent moves the square root to the denominator. So we have verified our answer.
To re-state the conclusion: given the differential equation \frac{dy}{dx} = \frac{x^2}{y}, the general antiderivative is y = \sqrt{\frac{2}{3} x^3 + C}.
Differential Equations with Initial Conditions:
You will, in general, have a C term at the end of most of the separable differential equations problems. Remember that this is because there are multiple functions that have the same derivative, and so there are many antiderivatives for one function.
You can, however, find a particular antiderivative, subject to an initial condition. That is, suppose you were given a small piece of additional information in the previous problem: your function goes through the point (3, 2).
So you have y = \sqrt{\frac{2}{3} x^3 + C}, and you know that x = 3, y = 2 satisfies this equation.
This allows you to solve for C.

y = \sqrt{\frac{2}{3} x^3 + C}

2 = \sqrt{\frac{2}{3} 3^3 + C}

2 = \sqrt{\frac{2}{3} 27 + C}

2 = \sqrt{18 + C}

2^2 = \left(\sqrt{18 + C}\right)^2

4 = 18+ C

C = -14

So the particular antiderivative of \frac{dy}{dx} = \frac{x^2}{y} that goes through the point (3, 2) is:

y = \sqrt{\frac{2}{3} x^3 -14}

Example 2:
Consider the differential equation \frac{dy}{dx} = (3x^2 - 4x)(y+2). Find the particular antiderivative such that y(2) = 10.
Separate:

\frac{dy}{dx} = (3x^2 - 4x)(y+2)

\frac{dy}{y+2} = (3x^2 - 4x) dx

Integrate:

\int \frac{dy}{y+2} = \int (3x^2 - 4x) dx

\int \frac{1}{y+2} dy= \int (3x^2 - 4x) dx

\ln|y+2| = 3 \frac{1}{3}x^3 - 4\frac{1}{2}x^2 + C

\ln|y+2| = x^3 - 2x^2 + C

Solve for y:

e^{\ln|y+2|} = e^{x^3 - 2x^2 + C}

|y+2| = e^{x^3 - 2x^2 + C}

|y+2| = e^{x^3 - 2x^2} e^C

(Recall that adding exponents is the same as multiplying two terms with the same base)

|y+2| = e^{x^3 - 2x^2} C = C e^{x^3 - 2x^2}

Replacing e^C with C is legitimate, since e^C is a constant. Don’t let the fact that they’re the same letter confuse you – we could call the new constant k or some other letter, but we’ll use the generic C because that is standard practice.

y+2 = \pm C e^{x^3 - 2x^2}

Note that the \pm in front of the C are not necessary, because C can be either positive or negative.

y = C e^{x^3 - 2x^2} - 2

If this problem did not have the initial condition $y(2) = 10$, we would be finished at this point, since we have that the general antiderivative of \frac{dy}{dx} = (3x^2 - 4x)(y+2) is y = C e^{x^3 - 2x^2} - 2. However, the initial condition allows us to find the specific antiderivative that passes through the point (2 , 10).
So we have:

y = C e^{x^3 - 2x^2} - 2

10 = C e^{2^3 - 2 \cdot 2^2} - 2

10 = C e^{8 - 8} - 2

10 = C e^0 - 2

10 = C\cdot 1 - 2

C = 12

And the particular antiderivative of \frac{dy}{dx} = (3x^2 - 4x)(y+2) satisfying the initial condition $y(2) = 10$ is:

y = 12 e^{x^3 - 2x^2} - 2

Final Notes:
Not all problems ask you to  solve for C, so it will frequently appear in your answer.
When you are asked to find C, it may not always be a nice integer like it was in these examples.
You may need to review a few of the integration rules, if you had trouble following anything in these examples.
As long as you follow the steps – Separate, Integrate, Solve for y – you should be alright!

 

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