Point) Consider the following differential equation: 2ye?xy +x+bxeexyy = 0.Find the value of b for which the equation is exact.(b) Solve the differentlal equation...

Match the differential equation with its solution.
Differential Equation $\quad$ Solution
$y^{\prime}-2 y=0 \quad$ (b) $y=-\frac{1}{2}+C e^{x^{2}}$

They asked us to verify that it satisfies the central vision for Let's Find the first year all of negatives he wanted for the negative to be for the X. We take our second riveted and we'll have C one e to the negative. We'll see to be to the X notices. The same thing someone has attracted him. One from me of it well, under 10 does satisfy the differential equation.

We want to solve a given differential equation. Dy dx plus Y X squared equals zero. Where we have initial condition, Y zero equals one. This question is challenging our ability to solve an initial value problem in particular is challenging us to use separation of variables to solve this iVP. There are four steps to execute. First will isolate X and Y on either side of the equation. Then we'll integrate both sides because they'll have a differential. Then we solve using integration techniques and finally we solve for our constant of integration using the ibP. So isolating X and Y gives whenever Y D y minus X squared dx. That's integrating has dy over Y equals negative X. 30 X. Or Ellen absolute value Y equals negative, execute over three plus C. We want to solve proceed to solve for the specific or particular solution. So we all get X equals zero, Y equals one L n Y L. In 10 equals zero cube over three plus C. This gives the equal zero. So our solution is Ellen absolute value Y equals negative X cubed over three

Hello, everyone. Today we're going to solve problem number 43 by a dash minus two weeks ago to zero by bicycle toe. Lex Integral divi equal toe integrate two weeks DX. So why you go to It's a square plus c thank you.

In this problem, we need to verify that the given differential equation is exact and then find the general solution. So let's start by identifying the functions M and N. So M if X and Y equals a squared minus two xy minus Y squared an N of X and Y is equal to negative X plus. Why? The whole squared now let's take the derivative of em when with respect to why? Yeah. So the derivative of a Square zero And the derivative of -2 X. Y is simply -2 x. And the derivative of Y squared is simply rather negative. Y squared is negative two. Y. Now let's take the derivative of end with respect to X. Or in setbacks equals. So we're using the power rule in the chain rule. So we get negative two times X plus Y. And we can subtract one from the power. So we get Race to the bar one times the derivative of the inside. Mhm. So the derivative of X plus Y with respect to X is just one We get and setbacks is equal to negative two works minus two Y. Now since M sub Y. And and subjects are equal. The differential equation is confirmed to be exact. Now all we need to do is find the general solution and for that will integrate. And on the right hand side we just get an arbitrary constant of integration. See kind of left hand side. We get a squared dance x minus two X squared times. Why divided by two minus X times Y squared minus Yeah X plus. Why? The whole cube Divided by three damaged the derivative of x plus y. It's just equal to one. We can simplify this a little bit and we get a ski a square times x minus x squared y. Another term X Y squared minus X. Plus. Why? The whole cube Divided by three equals c.