Differential Equations

Verify that is a solution to . (Note here that we use $y^{\iota }$to represent , and $y^{\iota \iota }$ to represent ).

Using the Product Rule, let us find the first and second derivative of y with respect to x.

Then

$y\text{'}\text{'}=\frac{d}{dx}(y\text{'})=\frac{d}{dx}(3e^{2x}+2x{e}^{2x})=6e^{2x}+2e^{2x}+4x{e}^{2x}=8e^{2x}+4x{e}^{2x}$

We can now fill in the values to get

Hence the solution is verified.

Find the general solution to.

This is of the form , so that means that the solution is of the form . This means we can fill in the two Functions to get to . Integrating the right-hand side, we get . On the left-hand side, this is a standard integral, given as .

This means that our solution is given as . Note we have combined both the constants into one here. We can simplify this further to give.

Find the solution to , with.

First, let us separate this equation to get .

The left-hand side integrates to , and the right-hand side integrates to.

These combine to give. This can simplify further to give.

We know that, so we can fill this in to get.

This gives C = 1, and our solution is.

Find the general solution to , and draw a graph showing this with four different particular solutions.

Below is a graph showing when C = -1, 0, 1, 2

A family of solutions, with C = 2 green, C = 1 blue, C = 0 red and C = -1 purple, Tom Maloy - StudySmarter Originals

Suppose there is a cylindrical water tank of radius 5m. The height of water in the tank at any point is denoted as. Water flows out of the tank at a rate proportional to the square root of the volume in the tank. Find .

The volume in the water tank at any one point is given as, meaning.

The water flows out at a rate proportional to the square root of the volume, meaning, where a is a constant of proportionality.

We can use the earlier expression for volume to give.

Then, by the Chain Rule,.