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 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.