The given differential equation is a linear differential equation of the form
+P(x)y=Q(x) where
P(x)== secx and
Q(x)==tanxThe integrating factor is given by:
I. F. =e∫P(x)dx=e∫ secxdx=elog| secx+tanx|= secx+tanx.
Multiplying both sides of the differential equation by the integrating factor, we get:
( secx+tanx)+( secx+tanx)y=( secx+tanx)tanx The left-hand side can be written as the derivative of the product of the integrating factor and the dependent variable, i.e.,
[( secx+tanx)y]=( secx+tanx)tanx.
Integrating both sides with respect to
x, we get:
( secx+tanx)y=∫( secx+tanx)tanxdxSimplifying the integral on the right-hand side:
∫( secx+tanx)tanxdx=∫( secxtanx+tan2x)dx=∫( secxtanx+ sec2x−1)dx= secx+tanx−x+C
Therefore, the general solution of the differential equation is:
( secx+tanx)y= secx+tanx−x+CUsing the initial condition
y(0)=1, we can find the value of the constant C :
( sec0+tan0)⋅1= sec0+tan0−0+C1=1+CC=0 Therefore, the particular solution of the differential equation is:
( secx+tanx)y= secx+tanx−xSo, the correct option is Option B.