Question

Find the solution of the differential equation that satisfies the given initial condition. y' tan x = 3a + y, y(π/3) = 3a, 0 &lt; x < π/2, where a is a constant.

Answer 1

Answer:

$y(x)=4a\sqrt{3}* sin(x)-3a$

Step-by-step explanation:

We have a separable equation, first let's rewrite the equation as:

$\frac{dy(x)}{dx} =\frac{3a+y}{tan(x)}$

But:

$\frac{1}{tan(x)} =cot(x)$

So:

$\frac{dy(x)}{dx} =cot(x)*(3a+y)$

Multiplying both sides by dx and dividing both sides by 3a+y:

$\frac{dy}{3a+y} =cot(x)dx$

Integrating both sides:

$\int\ \frac{dy}{3a+y} =\int\cot(x) \, dx$

Evaluating the integrals:

$log(3a+y)=log(sin(x))+C_1$

Where C1 is an arbitrary constant.

Solving for y:

$y(x)=-3a+e^{C_1} sin(x)$

$e^{C_1} =constant$

So:

$y(x)=C_1*sin(x)-3a$

Finally, let's evaluate the initial condition in order to find C1:

$y(\frac{\pi}{3} )=3a=C_1*sin(\frac{\pi}{3})-3a\\ 3a=C_1*\frac{\sqrt{3} }{2} -3a$

Solving for C1:

$C_1=4a\sqrt{3}$

Therefore:

$y(x)=4a\sqrt{3}* sin(x)-3a$