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Still Confused?

Try reviewing these fundamentals first.

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Try reviewing these fundamentals first.

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Get Started Now- Intro Lesson: a3:46
- Intro Lesson: b5:49
- Intro Lesson: c3:35
- Lesson: 1a1:14
- Lesson: 1b2:15
- Lesson: 1c2:25
- Lesson: 1d4:17
- Lesson: 2a2:17
- Lesson: 2b3:13
- Lesson: 2c4:17
- Lesson: 2d2:19
- Lesson: 3a2:11
- Lesson: 3b3:09
- Lesson: 46:13

In this lesson, we will look at the notation and highest order of differential equations. To find the highest order, all we look for is the function with the most derivatives. After, we will verify if the given solutions is an actual solution to the differential equations. We do this by simply using the solution to check if the left hand side of the equation is equal to the right hand side. Lastly, we will look at an advanced question which involves finding the solution of the differential equation.

We say that:

$y' (x) = \frac{dy}{dx}$ or $y'(t) = \frac{dy}{dt}$

Where:

1. $y'(x)$ is the first derivative of the function y in terms of $x$.

2. $y'(t)$ is the first derivative of the function y in terms of $t$.

$y' (x) = \frac{dy}{dx}$ or $y'(t) = \frac{dy}{dt}$

Where:

1. $y'(x)$ is the first derivative of the function y in terms of $x$.

2. $y'(t)$ is the first derivative of the function y in terms of $t$.

- IntroductionDifferential Equations Overviewa)Notation and Order of a Differential Equationb)Solution to a Differential Equationc)Particular Solutions
- 1.
**Finding the Order of a Differential Equation**

What is the order for the following differential equations?a)$y''' = {t^2} + 7$b)$y''( t ) + 6y'( t ) + 8y( t ) = ln( t )$c)$5\frac{{dy}}{{dt}} = \csc \left( {4t} \right) + \frac{{{d^2}y}}{{d{t^2}}}$d)${x^3}\frac{{{d^3}y}}{{d{x^3}}} + {x^2}\frac{{{d^2}y}}{{d{x^2}}} + x\frac{{dy}}{{dx}} = 1$ - 2.
**Verifying Solutions**

Show that the following functions is a solution to the differential equation:a)$y = \cos( 5t )$ is a solution to $y'' + 25y =$0b)$y = C\cos \left( {5t} \right)$ is a solution to $y'' + 25y =$0 where $C$ is a constantc)$y = C{e^x}$ is a solution to $3y + y' + y'' - 3y''' = 2C{e^x}$ where $C$ is a constantd)$y = \frac{{{x^4}}}{4} + {x^2}$ is a solution to $y''' = 6x$ - 3.
**Finding a Particular Solution**

You are given the general solution as well as the initial condition. Find the particular solution which suits the following initial conditions:a)$6{x^3} + 9{y^2} = C$ where $y\left( 0 \right) = 3$b)$C{e^{3t}} = {y^2}$ where $y\left( 1 \right) = 1$ - 4.
**Integrating to Find the General Solution**

Find the general solution of the differential equation$\;\frac{{dy}}{{dx}} = x{e^{{x^2}}}$.

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