Video: Solving Differential Equations

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

02:11

Video Transcript

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

Remember, we can find a general solution for the differential equation d𝑦 by dπ‘₯ equals eight π‘₯ plus three by first performing the reverse process for differentiation. We’re going to integrate our function for d𝑦 by dπ‘₯ with respect to π‘₯. So 𝑦 is equal to the indefinite integral of d𝑦 by dπ‘₯ with respect to π‘₯. Or 𝑦 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‘Žπ‘₯ to the 𝑛th power, where π‘Ž and 𝑛 are real constants and 𝑛 is not equal to negative one, is π‘Žπ‘₯ to the power of 𝑛 plus one over 𝑛 plus one plus some constant of integration 𝑐.

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 𝑐. So we obtained the general solution of our differential equation to be 𝑦 equals four π‘₯ squared plus three π‘₯ plus 𝑐.

Now, let’s refer back to the question. We’re told that 𝑦 of zero equals 12. In other words, at some point in the equation 𝑦 equals 𝑓 of π‘₯, when π‘₯ is equal to zero, 𝑦 is equal to 12. Let’s substitute these values into our equation and see what happens. We obtain 12 equals four times zero squared plus three times zero plus 𝑐. Well, this whole equation simplifies to 12 equals 𝑐. And that’s great because now we know the value of our constant. And we can, therefore, say that 𝑦 equals four π‘₯ squared plus three π‘₯ plus 12.

This is known as the particular solution. And 𝑦 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.

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