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Calculus

Higher order derivativesCalculus

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Order and solutions to differential equations- Home
- AP Calculus AB
- Differential Equations

Still Confused?

Try reviewing these fundamentals first.

Calculus

Higher order derivativesCalculus

AntiderivativesCalculus

Order and solutions to differential equationsStill Confused?

Try reviewing these fundamentals first.

Calculus

Higher order derivativesCalculus

AntiderivativesCalculus

Order and solutions to differential equationsNope, I got it.

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Get Started Now- Intro Lesson1:44
- Lesson: 1a7:00
- Lesson: 1b14:04
- Lesson: 1c6:24
- Lesson: 1d5:39
- Lesson: 1e4:39
- Lesson: 2a6:58
- Lesson: 2b9:10

Separable of variables is a method we use to find a general solution of a differential equation. The method involves separating all the y variables to the left hand side of the equation, and moving all the x variables to the right side. Afterwards, we integrate both sides of the equation and then isolate for y to find the general solution. If it is too hard to isolate for y, you can leave the answer as it is. We will also be looking at questions about particular solutions. These are solutions which involves finding the value of the constant, when given an initial value.

Basic concepts: Higher order derivatives, Antiderivatives, Order and solutions to differential equations,

A separable differential equation is in the following form:

$f(y)\frac{dy}{dx}=g(x)$

Where:

1. $f(x)$ is a function in terms of $y$.

2. $g(x)$ is a function in terms of $x$.

We want to convert the equation to the following form:

$f(y)dy=g(x)dx$

so that we can integral both sides, and solve for $y$.

$f(y)\frac{dy}{dx}=g(x)$

Where:

1. $f(x)$ is a function in terms of $y$.

2. $g(x)$ is a function in terms of $x$.

We want to convert the equation to the following form:

$f(y)dy=g(x)dx$

so that we can integral both sides, and solve for $y$.

- IntroductionSeparable Equations Overview
- 1.
**Separable Equations without Initial Conditions**

Find the general solution of the following differential equations:a)$\frac{4}{y^3}\frac{dy}{dx}=\frac{1}{x}$b)$\frac{dy}{dx}=(1+e^{-x})(y^2-1)$c)$\frac{dy}{dx}=\frac{(x+2)^2}{x \sin y}$d)$\frac{dy}{dx}=y(3+e^{2x})$e)$\frac{dy}{dx}=\frac{e^x}{y}$ - 2.
**Initial Value Problems**

Solve the following differential equations:a)$\frac{dy}{dx}=xy-x$ subject to $y(0)=2$b)$\frac{dy}{dx}=\frac{x(e^{x^2}+4)}{4y^2}$ subject to $y(0)=1$

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