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

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### 1 Answer

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.**