### Video Transcript

Find a particular solution for the
following differential equation for which π¦ of zero equals 12. dπ¦ by dπ₯ equals
eight π₯ plus three.

Remember, we can find a general
solution for the differential equation dπ¦ by dπ₯ equals eight π₯ plus three by
first performing the reverse process for differentiation. Weβre going to integrate our
function for dπ¦ by dπ₯ with respect to π₯. So π¦ is equal to the indefinite
integral of dπ¦ by dπ₯ with respect to π₯. Or π¦ equals the integral of eight
π₯ plus three with respect to π₯. We then recall that the integral of
a polynomial term of the form ππ₯ to the πth power, where π and π are real
constants and π is not equal to negative one, is ππ₯ to the power of π plus one
over π plus one plus some constant of integration π.

Essentially, we add one to the
exponent and then divide by that new number. This means the integral of eight π₯
is eight π₯ squared over two. Then the integral of three is three
π₯. And, of course, since this is an
indefinite integral, we end up with that constant of integration π. So we obtained the general solution
of our differential equation to be π¦ equals four π₯ squared plus three π₯ plus
π.

Now, letβs refer back to the
question. Weβre told that π¦ of zero equals
12. In other words, at some point in
the equation π¦ equals π of π₯, when π₯ is equal to zero, π¦ is equal to 12. Letβs substitute these values into
our equation and see what happens. We obtain 12 equals four times zero
squared plus three times zero plus π. Well, this whole equation
simplifies to 12 equals π. And thatβs great because now we
know the value of our constant. And we can, therefore, say that π¦
equals four π₯ squared plus three π₯ plus 12.

This is known as the particular
solution. And π¦ of zero equals 12 is an
initial value. These are sometimes called initial
value problems as they allow us to find a particular solution by giving us an output
of a specific π₯-value.