Let y = y(x) be solution of the differential equation [ _( dy dx ) = 3x + 4y], with y(0) =…

Question

Let y = y(x) be solution of the differential equation

$${\log _{}}\left( {{{dy} \over {dx}}} \right) = 3x + 4y$$, with y(0) = 0.

If $$y\left( { - {2 \over 3}{{\log }_e}2} \right) = \alpha {\log _e}2$$, then the value of $$\alpha$$ is equal to :

Accepted Answer

$${{dy} \over {dx}} = {e^{3x}}.{e^{4y}} \Rightarrow \int {{e^{ - 4y}}dy = \int {{e^{3x}}dx} } $$

$${{{e^{ - 4y}}} \over { - 4}} = {{{e^{3x}}} \over 3} + C \Rightarrow  - {1 \over 4} - {1 \over 3} = C \Rightarrow C =  - {7 \over {12}}$$

$${{{e^{ - 4y}}} \over { - 4}} = {{{e^{3x}}} \over 3} - {7 \over {12}} \Rightarrow {e^{ - 4y}} = {{4{e^{3x}} - 7} \over { - 3}}$$

$${e^{4y}} = {3 \over {7 - 4{e^{3x}}}} \Rightarrow 4y = \ln \left( {{3 \over {7 - 4{e^{3x}}}}} \right)$$

$$4y = \ln \left( {{3 \over 6}} \right)$$ when $$x =  - {2 \over 3}\ln 2$$

$$y = {1 \over 4}\ln \left( {{1 \over 2}} \right) =  - {1 \over 4}\ln 2$$

Solutions · 1