Question

Differential equation dy/dx= 3e^2x +4e^4x / e^x +e^-x

Answer 1

Answer:

this may be the answer.

Answer 2

Answer:

$y = \frac{4e^{3x}}{3} - e^x + tan^{-1}e^x + C$

Step-by-step explanation:

Given $\frac{dy}{dx} =\frac{3e^{2x} + 4e^{4x}}{e^{x} + e^{-x}}$ → (1)

Let $t=e^x$ → (2)

$dt = e^x dx$

$\frac{dt}{e^x} = dx$

$\frac{dt}{t} = dx$ → (3)

Using (2) and (3) in (1) we get

$dy = \frac{3t^2+ 4t^4}{t + \frac{1}{t} } \frac{dt}{t}$

$dy = \frac{3t^2+ 4t^4}{t^2 +1} dt$

$dy = (4t^2 - \frac{t^2}{t^2+1}) dt$

$dy = (4t^2 - \frac{t^2 +1 - 1}{t^2+1}) dt$

$dy = (4t^2 - (1 - \frac{1}{t^2+1})) dt$

$dy = (4t^2 - 1 + \frac{1}{t^2+1} ) dt$

Integrating both sides we get,

$\int dy = \int {(4t^2 - 1 + \frac{1}{t^2+1} }) \, dt$

$y = \frac{4t^3}{3} - t + tan^{-1}t + C$

Using (2) in the above equation, we get,

$y = \frac{4e^{3x}}{3} - e^x + tan^{-1}e^x + C$