Question

Show that $3e^x\tan ydx+(1-e^x){\sec }^{2}ydy=0$

Answer 1

For showing the above result, basically we have to find the solution of the given expression, for doing so we have to simplify the above equation in a such a way that function of x and y are form in either side of the equation and then integrate both sides to have the final solutions.

$3e^x\tan ydx+(1-e^x){\sec }^{2}ydy=0$....(1)

Now let consider

$e^{x-1}=t$ ,...(2)

$e^{x}dx=dt..$.(3)

puting 2 and 3 in (1)

$\frac{3dt}{t}=\frac{1}{{\cos }^{2}y\ast tany}dy$

$\frac{3dt}{t}=\frac{1}{{\cos }^{2}y\ast \frac{siny}{cosy}}dy$

$\frac{3dt}{t}=\frac{1}{\sin y\cos y}dy$

multiply and divide by 2

$\frac{3dt}{t}=\frac{2}{2\sin y\cos y}dy$

now we know that $2\sin y\cos y=2\sin 2y$

$\frac{3dt}{t}=\frac{2}{\sin 2y}dy$

$\frac{3dt}{t}=2\csc 2ydy$

$\int \frac{3dt}{t}=2\int \csc 2ydy$

integrate both sides,

$3\ln t=\ln (\csc 2y-\cot 2y)+c$.....(4)

putting the value of t in equation (4)

$3\ln \left(e^{x-1}\right)=\ln (\csc 2y-\cot 2y)+c$

This is the required solution for the given expression . and which eventually showed the required expression.