2. Show that each equation is homogeneous. Then, solve the...

2. Show that each equation is homogeneous. Then, solve the equation by substitution. (a) \frac{dy}{dx} = \frac{y - x}{y + x} (b) \frac{dy}{dx} = \frac{y^2 + yx}{x^2} (c) \frac{dy}{dx} = \frac{-y}{x(ln(x) - ln(y) - 1)}

Transcript

00:02 Okay, so there's several parts of this equation, this particular problem.

00:07 The first time you want to solve this differential equation, and we're asked if it's homogenous.

00:13 Well, the way to tell that is you write it out so that we can basically use algebra on the dx, and so our equation looks like this.

00:24 When we do that, so we multiply by the two, get all the denominators out of there.

00:30 And then every term has to have the same power of x and y in total.

00:37 So in this case, it's three.

00:40 Every term has somehow third power.

00:44 So then we know it's homogenous.

00:48 And then we make the substitution y is equal to ux.

00:52 So d y is x, du plus u d x, make that substitute that back in there.

00:57 Do a little rearranging of terms and we get that we can cancel out an x squared on both sides of the equation rearrange a little bit more bring the dxs over to the right there's a two there and i end up with an equation that looks like this after more rearranging i can integrate this easily.

01:27 I get the log of u squared plus one is log x plus log c, where c is some constant, or the u squared plus one is c squared x squared.

01:38 And i substitute u is equal to y over x.

01:42 I find out that y squared is x squared times c squared x squared minus one.

01:47 And that tells me basically y is a function of x.

01:50 A lot of times when we're solving these homogenous equations, we can't easily solve for one in terms of the other.

01:59 So this time we can, but a lot of times we can't.

02:05 In part b, we have two different equations we want to try and solve.

02:11 The first one looks like this with y of zero is equal to three.

02:15 So if i examine this equation carefully, i realize that i can take the y term and the x squared plus two, d, y, d, x, and i can combine them into a single derivative...

Question

2. Show that each equation is homogeneous. Then, solve the equation by substitution. (a) \frac{dy}{dx} = \frac{y - x}{y + x} (b) \frac{dy}{dx} = \frac{y^2 + yx}{x^2} (c) \frac{dy}{dx} = \frac{-y}{x(ln(x) - ln(y) - 1)}

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