# Video: Solving Differential Equations

Find a particular solution for the following differential equation for which π¦(0) = 12: dπ¦/dπ₯ = 8π₯ + 3.

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### 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. Weβre 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 ππ₯ to the πth power, where π and π are real constants and π is not equal to negative one, is ππ₯ 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, letβs refer back to the question. Weβre 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. Letβs 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 thatβs 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.