Find the general solution for each differential equation. Verify that each solution satisfies th

Find the general solution for each differential equation....

Key Concepts

Indefinite Integration

Indefinite integration is the process used to reverse differentiation. When solving a differential equation, once the derivative is isolated, integration is applied to obtain the original function plus an arbitrary constant, which represents the set of all antiderivatives.

Ordinary Differential Equations

This concept involves equations that relate a function and its derivatives. In solving such equations, the goal is often to find a function that satisfies the relationship between its derivatives and the independent variable.

General Solution

A general solution of a differential equation is a family of functions that includes all possible solutions, typically expressed with an arbitrary constant. This constant arises from the integration process, reflecting the fact that many functions can have the same derivative.

Verification of Solutions

Verification involves substituting the obtained general solution back into the original differential equation to ensure that it satisfies the equation. This process confirms the validity of the solution and demonstrates that no errors were made during integration or simplification.

Transcript

00:01 The differential equation, 4x cubed minus 2, d .y over dx equals zero.

00:05 And if i want to take this back to the general solution, i need to do some rewriting it first.

00:10 So i would actually solve it for d .y over dx, and then that way i can send the dx over and separate those variables.

00:17 So i would subtract that 4x cubed first, and then i would divide by the negative 2 so that i've isolated dy over dx.

00:30 When i divide by that negative 2, i should have positive 2x cubed on the right side by itself.

00:39 Now, if i multiply both sides by dx is effectively what i'm doing here.

00:44 I'm getting the differentials to be separated in terms of each variable.

00:49 On the left hand side, it's with respect to y, and on the right hand side with respect to x.

00:54 So now to take it from the differential form back to the original general solution, i want to integrate it.

01:01 That's the anti -derivative to the opposite of the differential.

01:07 And integrating d -y, it's like there's a one there.

01:10 You know how typically integral one would be x? well, here, since it's d -y, it's y.

01:16 And that's what gives me my y equals.

01:19 And then because i have an exponent of three, doing the anti -derivative takes it to an exponent of four, dividing by that new exponent would give me a one -half in front, right? a two over four...

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