00:01
In this question we have been given with the equation y' ' plus y ' tan x is equal to y ' the whole square.
00:11
We need to solve this second order differential equation.
00:15
So for that let assume that y ' is equal to v therefore v ' is equal to y' ' then the equation 1 becomes v ' plus v into tan x is equal to v square then dividing it by v square we will get v to the power minus 2 v ' plus 1 by v tan x is equal to 1.
00:47
Now take 1 by v is equal to u then differentiating we will get minus v to the power minus 2 v ' is equal to u ' which is equal to du by dx.
01:04
Then the equation will become minus du by dx plus 1 by v will be u, u into tan x is equal to 1.
01:17
Then du by dx minus tan x into u is equal to minus 1.
01:27
Let this be equation number 2.
01:29
Now equation 2 is the linear differential equation of first order.
01:35
So the integrating factor if is equal to e to the power minus integral tan x dx which is equal to e to the power log of modulus of cos x which is equal to cos x because exponential and logarithmic are inverse to each other they will become 1.
01:59
So multiply both sides of 2 by the integrating factor to make the exact differentiating.
02:06
Then we will get du by dx into cos x minus tan x u into cos x which is equal to minus 1 into cos x...