The only difference is: If you multiply or divide both sides of an equation by the same negative number, the equation remains the same, but If you multiply or divide both sides of an inequality by the same negative number, the inequality reverses.

5 0

3 years ago

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

Paladinen [302]

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:

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

3 0

4 years ago