## Separation Of Variables

### Video List

A01  separable 1

A02  ivp 1

A03  ivp 2

PatrickJMT Videos

A05  separable 1

A06  separable 2

A07  separable 3

A04  particular 1

A08  particular 2

A09  particular 3

A10  separable 1

A11  separable 2

A12  separable 3

A13  separable 4

A16  separable 5

A14  B02  separable 1

A15  separable 2

classifications

first-order

separable

This is one of the simplest techniques when solving differential equations and sometimes it is introduced in first year calculus while studying integration. This technique is used when it is possible to isolate the variables, using algebra and differentials, on separate sides of the equal sign. This allows us to integrate both sides of the equation individually with only one variable on each side. An equation that allows us to apply this technique is called a separable differential equation.

The idea with this technique is that the differential equation is in a form where we can isolate the two variables to each side of the equal sign. Then we can integrate each side separately. It's really that easy. The trick is to use algebra to get the equation into the right form. Also, some instructors (most, from what I've seen), do a trick with constants that confuse a lot of people. So, we will go through that so that, if your instructor does it, you will know why. Let's do an example to demonstrate the technique.

Example
This simple example will demonstrate the technique.
Find $$y$$ for $$y' = x$$
First, we rewrite the derivative so that it is more obvious what we need to do.
$$\displaystyle{ y' = x ~~~ \to ~~~ \frac{dy}{dx} = x }$$
Now we write the differential equation by 'moving' the $$dx$$ to the other side. It looks like we are multiplying $$dx$$ on both sides but that's not what is really happening. However, it helps me remember what to do by thinking of it this way. What we are doing is writing the equation in differential form. So now we have
$$\displaystyle{ \frac{dy}{dx} = x ~~~ \to ~~~ dy = x~dx}$$
Now we integrate both sides. On the left we integrate with respect to $$y$$, on the right with respect to $$x$$.
$$\displaystyle{ \begin{array}{rcl} \int{dy} & = & \int{x~dx} \\ y + c_1 & = & \frac{x^2}{2} + c_2 \end{array} }$$
Okay, so now we have two constants, one from each integration (don't forget these constants; they are extremely important when solving differential equations). We are going to move $$c_1$$ to the other side and combine it with $$c_2$$.
$$\displaystyle{ y = \frac{x^2}{2} + c_2-c_1 }$$
Let $$C = c_2 - c_1$$. We then rewrite this as
$$\displaystyle{ y = \frac{x^2}{2} + C }$$ and we are done.

What many instructors do is skip the steps showing the two constants and combining them, i.e. $$C = c_2 - c_1$$ . They just jump to one constant. This can be confusing at first, if you don't know what is happening.

Here is an in-depth video discussing first-order linear equations, separation of variables and steady-state and transient solutions.   open video

Next - - After you work through the practice problems below, your next topic is linear, first-order differential equations.

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### Search 17calculus

Separation Of Variables Practice Problems

Instructions - - Unless otherwise instructed, solve these separable differential equations and find the particular solution, if possible, giving your answers in exact terms.

 Level A - Basic

Practice A01

$$\displaystyle{ \frac{dy}{dx} = xy }$$

Practice A02

$$\displaystyle{ \frac{dy}{dx} = y^2(1+x^2) }$$ $$y(0) = 1$$

Practice A03

$$\displaystyle{ y \frac{dy}{dx} = x^2 + sech^2(x) }$$ $$y(0) = 2$$

Practice A04

$$\displaystyle{ \frac{dy}{dx} = cos(x) }$$; $$y(0) = -1$$

Practice Problem [ A01 ] Solution

$$\displaystyle{ y = Ae^{x^2/2} }$$

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Practice Problem [ A02 ] Solution

$$\displaystyle{ y(x) = \frac{-1}{x^3/3 + x -1} }$$

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Practice Problem [ A03 ] Solution

$$\displaystyle{ y^2 = 2 [ x^3/3 + tanh(x) ] + x + 4 }$$

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Practice Problem [ A04 ] Solution

$$\displaystyle{ y=\sin(x)-1 }$$

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Practice A05

$$\displaystyle{ \frac{dy}{dx} = x/y^2 }$$

Practice A06

$$\displaystyle{ \frac{dy}{dx} \cdot \frac{y^3+y}{x^2+3x} = 1 }$$

Practice A07

$$\displaystyle{ \frac{dy}{dx} = \frac{\cos(x)}{y-1} }$$

Practice A08

$$\displaystyle{ \frac{dy}{dx} = e^{4x-y} }$$; $$y(0) = 5$$

Practice Problem [ A05 ] Solution

$$\displaystyle{ y = \sqrt[3]{3x^2/2 + C} }$$

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Practice Problem [ A06 ] Solution

$$\displaystyle{ \frac{y^4}{4} + \frac{y^2}{2} = \frac{x^3}{3} + \frac{3x^2}{2} + C }$$

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Practice Problem [ A07 ] Solution

$$\displaystyle{ \frac{y^2}{2} - y = \sin(x) + C }$$

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Practice Problem [ A08 ] Solution

$$\displaystyle{ y = ln\left[ \frac{e^{4x}}{4} + e^5 - \frac{1}{4} \right] }$$

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Practice A09

$$\displaystyle{ \frac{y+2}{x^2-x+2} \frac{dy}{dx} = \frac{x}{y} }$$ $$y(1) = 2$$

Practice A10

$$\displaystyle{ \frac{dy}{dx} = 2x\sqrt{y-1} }$$

Practice A11

$$\displaystyle{ \frac{dy}{dx} = \frac{x^2+1}{x^2(3y^2+1)} }$$

Practice A12

$$\displaystyle{ \frac{dy}{dx} = \frac{4-2x}{3y^2-5} }$$; $$y(1) = 3$$

Practice Problem [ A09 ] Solution

$$\displaystyle{ \frac{y^3}{3} + y^2 = \frac{x^4}{4} - \frac{x^3}{3} +x^2 + 23/4 }$$

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Practice Problem [ A10 ] Solution

$$\displaystyle{ y = 1 + 1/4(x^2+C)^2 }$$

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Practice Problem [ A11 ] Solution

$$\displaystyle{ y^3 + y = x - 1/x + C }$$

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Practice Problem [ A12 ] Solution

$$\displaystyle{ y^3 - 5y = 4x-x^2 + 9 }$$

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Practice A13

$$\displaystyle{ \frac{dy}{dx} = 2\sqrt{y} }$$; $$y(0) = 9$$

Practice A14

$$\displaystyle{ \frac{dy}{dx} = \frac{x^2}{1-y^2} }$$

Practice A15

$$\displaystyle{ \frac{dy}{dx} = \frac{y \cos(x)}{1+2y^2} }$$; $$y(0)=1$$

Practice A16

$$\displaystyle{ \frac{du}{dt} = \frac{2t+sec^2(t)}{2u} }$$; $$u(0) = -5$$

Practice Problem [ A13 ] Solution

$$\displaystyle{ y = (x+3)^2 }$$

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Practice Problem [ A14 ] Solution

$$\displaystyle{ 3y-y^3 = x^3 + C }$$

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Practice Problem [ A15 ] Solution

$$\displaystyle{ ln|y| + y^2 = \sin(x) + 1 }$$

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Practice Problem [ A16 ] Solution

$$\displaystyle{ u^2 = t^2 + \tan(t) + 25 }$$

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 Level B - Intermediate

Practice B01

$$\displaystyle{ dv/ds = (s+1) / (sv+s) }$$

Practice B02

$$\displaystyle{ \frac{dy}{dx} = \frac{3x^2 + 4x+2}{2(y-1)} }$$ $$y(0)=-1$$

Practice Problem [ B01 ] Solution

From the $$dv/ds$$ term, we can tell that $$s$$ is the independent variable and $$v$$ is the dependent variable, so that the solution will be $$v(s)$$. Let's try separation of variables.

Most of this should be easy to follow. There just a few clarifications in order.
1. We can combine all constants under the square root into one by letting $$C=1+c_1$$. This is not required but cleans up the equation.
2. Adding $$1$$ to each side after integrating allows us to complete the square.
4. When taking the square root, don't assume your answer will be positive. Use $$\pm$$ unless you can show that the result will always be positive (or negative).
3. Depending on what your instructor requires, you may not need to solve for $$v$$. Check to see what they require.

$$\displaystyle{ \begin{array}{rcl} dv/ds & = & \frac{(s+1)}{(sv+s)} \\ & = & \frac{(s+1)}{s(v+1)} \\ (v+1)\frac{dv}{ds} & = & \frac{(s+1)}{s} \\ (v+1) dv & = & \frac{(s+1)}{s} ds \\ \int{v+1 ~dv} & = & \int{ \frac{s+1}{s} ds } \\ & = & \int{ 1 + \frac{1}{s} ds } \\ \frac{v^2}{2} + v & = & s + ln(s) + c_1 \\ v^2 + 2v & = & 2s + 2ln(s) + c_1 \\ v^2 + 2v + 1 & = & 2s + ln(s) + 1 + c_1 \\ (v+1)^2 & = & \\ v & = & \pm \sqrt{2s + ln(s) + 1 + c_1} - 1 \\ v & = & \pm \sqrt{2s + ln(s) + C} - 1 \end{array} }$$

$$\displaystyle{ v = \pm \sqrt{2s + ln(s) + C} - 1 }$$

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Practice Problem [ B02 ] Solution

$$\displaystyle{ y = 1-\sqrt{x^3+2x^2+2x+4) }$$