21–32. Finding general solutions Find the general solution of ...

Question

21–32. Finding general solutions Find the general solution of each differential equation. Use C,C1,C2... to denote arbitrary constants.

y'(t) = t lnt + 1

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says to determine the general solution of the differential equation, and we're given that F of U is equal to U, multiplied by the natural log of u squared plus 3. The first thing we're going to do is rewrite our differential equation. So we're going to rewrite this as the derivative of F with respect to U is equal to, and here we're going to rewrite the U multiplied by the natural log of U2 using the properties logarithm. So this is going to be equal to 2 U, multiplied by the natural log of the absolute value of U. and then we'll still have R + 3. The next we'll multiply both sides of the equation by DU doing so, we'll have that DF is going to be equal to. The quantity of 2 you multiplied by the natural log of the absolute value of U plus 3 in quantity DU. And now we just need to integrate both sides of this equation. And so, in a gradi, we'll get that F of you is going to be equal to. The integral of tou, multiplied by the natural log of the absolute value of U D U. Plus the interval of 3 DU. So here we've just taken our integral on the right hand side and broken it up into two separate integrals, and we're going to evaluate each one of these integrals independently. So for the first integral that we're going to look at is going to be the integral of 2U multiplied by the natural log of the absolute value of UDU. And to integrate this, we're going to need to integrate by parts. So to integrate by parts, we'll set W is equal to the natural log of the absolute value of U. And that means that DV is going to be equal to 2UDU. So, that also means that DW is going to be equal to DU divided by U and V is going to be equal to U squared. So, integrating web parts are integral is going to be equal to, and here we call when you integrate web parts, the first term is going to be V multiplied by W, which is going to be U squared, multiplied by the natural log of the absolute value of U. And then we'll have minus the integral of VDW, which is going to be U squared multiplied by DU divided by U. And so we can simplify that integra. That means our integral here is going to be equal to. The first term remains the same. So we'll have you squared multiplied by the natural log of the absolute value of U, and we'll have minus the integral of UDU, which we can integrate. So integrating that we'll have this is equal to the first term again remains the same of U squared multiplied by the natural log of the absolute value of U. Minus, and here when we integrate you with respect to you, we get 1/2 multiplied by U squared, and then we'll have to add an integration constant of plus C. Next, we'll look at the 2nd interval in our function F of you. And for that integral, we'll have the integral of 3DU, which we can integrate directly, and this is going to be equal to 3U plus C. So now we can combine our two results together, and we'll see that F of you. It's going to be equal to. U squared multiplied by the natural log of the absolute value of U. Minus 1/2 multiplied by U2. Plus 3 U plus C. So here we've combined our two integration constants into a single integration constant of plus C. And this here is the answer that we were looking for for this problem. So I hope that this explanation has been useful, and I'll see you in the next video.

21–32. Finding general solutions Find the general solution of each differential equation. Use C,C1,C2... to denote arbitrary constants.y'(t) = t lnt + 1

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21–32. Finding general solutions Find the general solution of each differential equation. Use C,C1,C2... to denote arbitrary constants.
y'(t) = t lnt + 1