Question

Find the particular solution of the differential equation ${e^x}\,\tan \,y\,\,dx\, + \,(2 - {e^x})\,{\sec ^2}y\,dy = 0$ , given that $y = \dfrac{\pi }{4}$ when $x = 0$

Answer 1

According to the given equation, first rearrange all the x terms on one side and y terms on the other side. Then integrate both the left and right side separately by substitution method that is $t = 2 - {e^x}$ and $\tan \,y = u$ .Then substitute the given values of x and y in the calculated integral equation. At last we get an equation that is the particular solution of the differential equation.

Complete step-by-step solution:
As it is given that, $y = \,\dfrac{\pi }{4}$ when $x = 0$ And we have to find the particular solution of the differential equation that is ${e^x}\tan \,y\,dx\, + \,(2 - {e^x})\,\,{\sec ^2}y.\,dy = 0$
Taking $(2 - {e^x})\,\,{\sec ^2}y.\,dy$ on right side of the equation,
So, we get
${e^x}\,\tan \,y\,dx = \, - (2 - \,{e^x}).\,{\sec ^2}y\,dy$
Re-shifting everything and taking integration on both side we get,
$\Rightarrow \,\int { - \dfrac{{{e^x}}}{{2 - {e^x}}}\,dx\,\, = \,\,} \,\int {\dfrac{{{{\sec }^2}y}}{{\tan \,y}}\,dy}$ eq. (1)
Now, take $t = 2 - {e^x}$
Therefore, on differentiating $t$ with respect to $x$ we get,
$\dfrac{{dt}}{{dx}} = \, - {e^x}$
Taking dx on the right side,
$\therefore \,dt = \, - {e^x}\,dx$ eq. (2)
Now, for the right-hand side we have
$\tan \,y = u$
Now differentiating with respect to $y$ we get,
${\sec ^2}y = \dfrac{{du}}{{dx}}$
Calculating the value of du in terms of dx
$\Rightarrow \,du = {\sec ^2}y.\,dx$ eq. (3)
Now, putting the value of eq. (1) and eq. (2) in eq. (1) we get,
$\int {\dfrac{{dt}}{t}} = \,\int {\dfrac{1}{u}\,du}$ [Remember $\int {\dfrac{1}{x}dx} = \,\ln \,x$ ]
So, we get
$\Rightarrow \,\ln \,\,t = \,\ln \,u + \,c$
By substituting the value of $u$and $t$ we get,
$\Rightarrow \ln (2 - {e^x})\, = \,\ln \,\tan \,y + c$
or
$\Rightarrow \ln (2 - {e^x})\, = \,\ln (\tan \,y)\, + c$
Now putting the value of $y = \dfrac{\pi }{4}$ and $x = 0$ we get
$\ln (2 - {e^0})\, = \ln (\tan \,\dfrac{\pi }{4})\, + \,c$
As, we know ${e^0} = 1$ and $\tan \dfrac{\pi }{4} = 1$
$\Rightarrow \ln (2 - 1) = \ln (1) + c\,$
And we also know that $\ln \,1 = 0$
$\Rightarrow \,0 = 0 + c$
So, we get the value of c
$\Rightarrow \,c = 0$
Therefore, the value of c is zero.
Now, putting the value of c in the given equation we get,
$\ln (2 - {e^x}) = \,\ln (\tan \,y)\, + 0$
$\Rightarrow \ln (2 - {e^x}) = \ln (y)$
Taking $\ln (y)$ on the left side of the equation we get,
$\Rightarrow \ln (2 - {e^x})\, - \,\ln (y) = 0$
As we know $\ln \left( A \right) - \ln \left( B \right) = \ln \left( {\dfrac{A}{B}} \right)$
$\Rightarrow \ln \left( {\dfrac{{2 - {e^x}}}{y}} \right) = 0$
Taking ln on the light side which becomes e so we get,
$\Rightarrow \,\dfrac{{2 - {e^x}}}{y} = {e^0}$
As we know ${e^0} = 1$
$\Rightarrow \,\dfrac{{2 - {e^x}}}{y} = 1$
Takin y on left side so,
\Rightarrow 2 - {e^x} = y$$$$ \Rightarrow {e^x} + y = 2
On rearranging the terms we get,
$\Rightarrow {e^x} + y - 2 = 0$

Thus, the particular solution of the differential equation is
${e^x} + y - 2 = 0$

Note: To solve these types of questions, do not forget to separate the variable on the left and right side of the equation and use substitution method for integration in a simpler way as done in the given question. Remember all the values of ${e^0} = 1$ , $\ln \,1 = 0$ and $\tan \dfrac{\pi }{4} = 1$ to get the required answer.