# Solve the differential equation `dy/(d theta) = (e^y * sin^2(theta))/(ysec(theta))`

This equation can be solved by the separation of variables. This means, we can rewrite this equation in a way that all y-containing terms are on the left side and all theta-containing terms are on the right side.

To do this, we can multiply both sides of equation by y...

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

This equation can be solved by the separation of variables. This means, we can rewrite this equation in a way that all y-containing terms are on the left side and all theta-containing terms are on the right side.

To do this, we can multiply both sides of equation by y and `d(theta)` :

`ydy = e^y * (sin^2(theta))/sec(theta) d(theta)`

We can also divide both sides of the equation by `e^y` :

`(ydy)/e^y = (sin^2(theta))/sec(theta) d(theta)`

Next, we can simplify both sides of the equation using the rule of exponents:

`1/e^y = e^(-y)`

and the reciprocal identity: `1/sec(theta) = cos(theta)` .

`ye^(-y)dy = sin^2(theta)cos(theta)d(theta)`

Now we can take the integral of the both sides of the equations.

`int (ye^(-y)dy) = int (sin^2(theta)cos(theta)d(theta))`

On the left side, we have a product of an exponential function and a polynomial function (y), so this integral will have to be taken by parts.

Let u = y and `dv = e^(-y)dy`

Then, du = dy and `v = int e^(-y)dy = -e^(-y)`

According to the integration by parts procedure,

`int udv = uv - int (vdu)`

Thus, `int (ye^(-y)dy = -ye^(-y) - int (-e^(-y))dy = -ye^(-y) - e^(-y)`

This will be the left side of the equation after the integration.

On the right side, we have the trigonometric integral which can be solved by substitution. Let `u = sin(theta)` . Then, `du = cos(theta)d(theta)` .

Plugging this into the integral, we get

`int sin^2(theta)cos(theta)d(theta) = int u^2 * du`

This is a power function, integration of which results in `u^3/3 + C` , where C is an arbitrary constant.

Substituting the expression for u in terms of theta back, we see that the left side after the integration becomes

`1/3 sin^3(theta) + C`

So, the equation is now

`-ye^(-y) - e^(-y) = 1/4 sin^3(theta) + C`

We cannot express `y(theta) ` explicitely from here, but we can solve for `theta(y)` . First, isolate `sin^4(theta)` :

`sin^3(theta) = 3(-ye^(-y) - e^(-y) - C)`

Than, take 4th degree root:

`sin(theta) = root(4) (-4(ye^(-y) + e^(-y) + C))`

Finally, take the inverse sine of both sides:

`theta = arcsin(root(3) (-3(ye^(-y) + e^(-y) + C)))`

This is the solution of the given equation. Note that it will be defined, depending on the value of constant C, only for restricted values of y.

Approved by eNotes Editorial Team