JEE MAIN - Mathematics (2020 - 4th September Evening Slot - No. 18)
The solution of the differential equation
$${{dy} \over {dx}} - {{y + 3x} \over {{{\log }_e}\left( {y + 3x} \right)}} + 3 = 0$$ is:
(where c is a constant of integration)
$${{dy} \over {dx}} - {{y + 3x} \over {{{\log }_e}\left( {y + 3x} \right)}} + 3 = 0$$ is:
(where c is a constant of integration)
$$x - {1 \over 2}{\left( {{{\log }_e}\left( {y + 3x} \right)} \right)^2} = C$$
$$y + 3x - {1 \over 2}{\left( {{{\log }_e}x} \right)^2} = C$$
x – loge(y+3x) = C
x – 2loge(y+3x) = C
Explanation
$${{dy} \over {dx}} - {{y + 3x} \over {\ln (y + 3x)}} + 3 = 0$$
Let $$\ln (y + 3x) = t$$
$${1 \over {y + 3x}}.\left( {{{dy} \over {dx}} + 3} \right) = {{dt} \over {dx}}$$
$$ \Rightarrow \left( {{{dy} \over {dx}} + 3} \right) = {{y + 3x} \over {\ln (y + 3x)}}$$
$$ \therefore $$ $$\left( {y + 3x} \right){{dt} \over {dx}} = {{y + 3x} \over t}$$
$$ \Rightarrow tdt = dx$$
$${{{t^2}} \over 2} = x + c$$
$${1 \over 2}{\left( {\ln (y + 3x)} \right)^2} = x + c$$
Let $$\ln (y + 3x) = t$$
$${1 \over {y + 3x}}.\left( {{{dy} \over {dx}} + 3} \right) = {{dt} \over {dx}}$$
$$ \Rightarrow \left( {{{dy} \over {dx}} + 3} \right) = {{y + 3x} \over {\ln (y + 3x)}}$$
$$ \therefore $$ $$\left( {y + 3x} \right){{dt} \over {dx}} = {{y + 3x} \over t}$$
$$ \Rightarrow tdt = dx$$
$${{{t^2}} \over 2} = x + c$$
$${1 \over 2}{\left( {\ln (y + 3x)} \right)^2} = x + c$$
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