To find the special solution of the differential equation y '' = e ^ (2Y) when x = 0, y = y '= 0; to write down the additional points of the steps

To find the special solution of the differential equation y '' = e ^ (2Y) when x = 0, y = y '= 0; to write down the additional points of the steps

As follows:
There is no X-type
Let y '= P, Y "= PDP / dy
The original differential equation is
pdp/dy=e^(2y)
That is, PDP = e ^ (2Y) dy
Two side integral
∫pdp=∫e^(2y)dy
We obtain P & sup2; = e ^ (2Y) + C '
Under the initial conditions x = 0, y = y '= 0, C' = - 1 is obtained
p=±√[e^(2y)-1]=dy/dx
Separate variables
dy/√[e^(2y)-1]=±dx
Make up the differential
1/√[1-e^(-2y)]d(e^-y)=±dx
Two sides integral
arcsine^(-y)=±x+C"
The initial conditions are x = 0, y = y '= 0
So C "= π / 2
So the special solution of the differential equation is
arcsine^(-y)=±x+π/2
Or sin (± x + π / 2) = e ^ (- y); cosx = e ^ (- y)