Question

#JEE
$\huge \red {\tt{ \frac{dy}{dx} + {x}^{2} = {x}^{2} . {e}^{3y} }}$
Using Variable separable ​

Answer 1

Step-by-step explanation:

We have,

$\rm{ \frac{dy}{dx} + {x}^{2} = {x}^{2} . {e}^{3y} }$

$\rm { \implies \frac{dy}{dx} = {x}^{2} . {e}^{3y} - {x}^{2} }$

$\rm { \implies \frac{dy}{dx} = {x}^{2} ({e}^{3y} - 1) }$

$\rm { \implies \frac{dy}{({e}^{3y} - 1)} = {x}^{2} dx }$

$\rm { \implies \int\frac{dy}{{e}^{3y} - 1} = \int{x}^{2} dx }$

$\rm { \implies \int\frac{(1 - {e}^{3y} + {e}^{3y}) dy}{{e}^{3y} - 1} = \int{x}^{2} dx }$

$\rm { \implies \int\frac{ {e}^{3y}}{{e}^{3y} - 1} dy - \int \frac{ {e}^{3y} - 1 }{ {e}^{3y} - 1 } dy = \frac{{x}^{3}}{3} + C }$

$\rm { \implies \int\frac{ {e}^{3y}}{{e}^{3y} - 1} dy - \int dy = \frac{{x}^{3}}{3} + C }$

$\rm { \implies ln({e}^{3y} - 1) - y = \frac{{x}^{3}}{3} + C }$

$\rm { \implies ln({e}^{3y} - 1) - ln \: ({e}^{ y} )= \frac{{x}^{3}}{3} + C }$

$\rm { \implies ln \bigg( \frac{{e}^{3y} - 1}{ {e}^{ y} } \bigg)= \frac{{x}^{3}}{3} + C }$

$\rm { \implies \frac{{e}^{3y} - 1}{ {e}^{ y} } = e^{ \frac{{x}^{3}}{3} + C }}$

$\rm { \implies \frac{{e}^{3y} - 1}{ {e}^{ y} } = e^{ \frac{{x}^{3}}{3} }. {e}^{C} }$

$\rm { \implies \frac{{e}^{3y} - 1}{ {e}^{ y} } = k e^{ \frac{{x}^{3}}{3} } } \: \: \: ....(where \: \: k = {e}^{ C} )$

$\rm { \implies {e}^{2y} - {e}^{ - y} = k e^{ \frac{{x}^{3}}{3} } }$

Answer 2

$\large\underline{\sf{Solution-}}$

Given Differential equation is

$\red{\rm :\longmapsto\:\tt{ \dfrac{dy}{dx} + {x}^{2} = {x}^{2} . {e}^{3y}}}$

can be rewritten as

$\rm :\longmapsto\:\tt\dfrac{dy}{dx} = {x}^{2} . {e}^{3y} - {x}^{2}$

$\rm :\longmapsto\:\tt\dfrac{dy}{dx} = {x}^{2}({e}^{3y} -1)$

On separate the variables, we get

$\rm :\longmapsto\:\tt\dfrac{dy}{ {e}^{3y} - 1} = {x}^{2} \: dx$

Now, on integrating both sides, we get

$\rm :\longmapsto\:\tt\displaystyle\int\tt \dfrac{dy}{ {e}^{3y} - 1} = \displaystyle\int\tt {x}^{2} \: dx$

We know,

$\boxed{ \rm{ \displaystyle\int\tt {x}^{n} \: dx \: = \frac{ {x}^{n + 1} }{n + 1} + c}}$

Using this identity, we get

$\rm :\longmapsto\:\tt\displaystyle\int\tt \dfrac{1 + {e}^{3y} - {e}^{3y} }{ {e}^{3y} - 1} \: dy= \frac{ {x}^{2 + 1} }{2 + 1} + c$

$\rm :\longmapsto\:\tt\displaystyle\int\tt \dfrac{{e}^{3y} -( {e}^{3y} - 1)}{ {e}^{3y} - 1} \: dy= \frac{ {x}^{3} }{3} + c$

$\rm :\longmapsto\:\tt\displaystyle\int\tt \dfrac{{e}^{3y}}{ {e}^{3y} - 1} \: dy - \displaystyle\int\tt \frac{{e}^{3y} - 1}{{e}^{3y} - 1} dy= \frac{ {x}^{3} }{3} + c$

$\rm :\longmapsto\: \dfrac{1}{3} \tt\displaystyle\int\tt \dfrac{3{e}^{3y}}{ {e}^{3y} - 1} \: dy - \displaystyle\int\tt 1 dy= \frac{ {x}^{3} }{3} + c$

We know,

$\boxed{ \rm{ \displaystyle\int\tt kdx = kx + c}}$

and

$\boxed{ \rm{ \displaystyle\int\tt \frac{f'(x)}{f(x)} \: dx \: = \: log |f(x)| + c}}$

So, using these, we get

$\rm :\longmapsto\:\dfrac{1}{3}log |{e}^{3y} - 1| - y = \dfrac{ {x}^{3} }{3} + c$

can be further rewritten as,

$\rm :\longmapsto\:log |{e}^{3y} - 1| - 3y = {x}^{3} +3 c$

$\bf :\longmapsto\:log |{e}^{3y} - 1| - 3y = {x}^{3} +d \: where \: d \: = \: 3c$