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

Hint: 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.