Let y=y(x) be solution of the differential equation log_e( dy dx)=3x+4y, with y(0)=0. If y(- 2 3 log_e2)=α log_e2, then the value of α is equal to:

Given,

$${\mathrm{log}}_{e}(\frac{dy}{dx})=3x+4y$$loge​(dxdy​)=3x+4y

$$\Rightarrow \frac{dy}{dx}={e}^{3x}\cdot {e}^{4y}$$⇒dxdy​=e3x⋅e4y

$$\Rightarrow \int {e}^{-4y}dy=\int {e}^{3x}dx$$⇒∫e−4ydy=∫e3xdx

$$\Rightarrow \frac{{e}^{-4y}}{-4}=\frac{{e}^{3x}}{3}+C$$⇒−4e−4y​=3e3x​+C

Given,

$$y(0)=0$$y(0)=0

So,

$$-\frac{1}{4}-\frac{1}{3}=C\Rightarrow C=-\frac{7}{12}$$−41​−31​=C⇒C=−127​

So, the particular solution is

$$\frac{{e}^{-4y}}{-4}=\frac{{e}^{3x}}{3}-\frac{7}{12}$$−4e−4y​=3e3x​−127​

$$\Rightarrow {e}^{-4y}=\frac{4{e}^{3x}-7}{-3}$$⇒e−4y=−34e3x−7​

$$\Rightarrow {e}^{4y}=\frac{3}{7-4{e}^{3x}}\Rightarrow 4y=\mathrm{ln}(\frac{3}{7-4{e}^{3x}})$$⇒e4y=7−4e3x3​⇒4y=ln(7−4e3x3​)

$4y=\ell n(\frac{3}{6})$4y=ℓn(63​) when $x=-\frac{2}{3}\ell n2$x=−32​ℓn2

$$\Rightarrow y=\frac{1}{4}\ell n(\frac{1}{2})$$⇒y=41​ℓn(21​)

$$\Rightarrow y=-\frac{1}{4}\ell n2$$⇒y=−41​ℓn2

So, $\alpha =-\frac{1}{4}$α=−41​