solve cos x(e^(2y) - y ) dy-e^y sin2x dx =0 with the condition y(0)=0
Answers
Step-by-step explanation:
Correct option is
A
199
siny
dx
dy
=cosx(2cosy−sin
2
x) ...(1)
Let R(x,y)=−cosx(2cosy−sin
2
x) and S(x,y)=siny
This is not an exact equation, because
dy
d(R(x,y))
=2cosxsiny
=0=
dx
d(S(x,y))
Therefore taking integrating factor u=u(x) above become exact.
⇒
dy
d(uR(x,y))
=
dx
d(uS(x,y))
⇒2sinycosxu=siny
dx
du
⇒
u
dx
du
=2cosx
Integrating both sides w.r.t x
logu=2sinx⇒u=e
2sinx
Multiplying both sides of (1) by u, we get
Let P(x,y)=−e
2sinx
cosx(2cosy−sin
2
x)
and Q(x,y)=e
2sinx
siny
Define f(x,y) such that
dx
d(f(x,y))
=P(x,y) and
dy
d(f(x,y))
=Q(x,y)
Then, the solution will be given by f(x,y)=c where c is constant
Now,
Integrating
dx
d(f(x,y))
w.r.t x in order to find f(x,y)
=−
4
1
(e
2sinx
(cosx+4cosy+2sinx−2))+g(y)
where g(y) is an arbitrary function of y.
Differentiating f(x,y) w.r.t to y in order to find g(y)
dx
d(f(x,y))
=
dy
d
(−
4
1
(e
2sinx
(cosx+4cosy+2sinx−2))+g(y))
=sinye
2sinx
+
dy
dg(y)
Substituting
dx
d(f(x,y))
in Q(x,y)
sinye
2sinx
+
dy
dg(y)
=sinye
2sinx
⇒
dy
dg(y)
=0
⇒g(y)=0
Substituting g(y) in f(x,y)
f(x,y)=−
4
1
(e
2sinx
(cosx+4cosy+2sinx−2))
Then solution is
−
4
1
(e
2sinx
(cosx+4cosy+2sinx−2))=c
⇒y=cos
−1
(
4
1
(ce
2sinx
−2sinx−cos2x+2))
For y(0)=
2
π
⇒c=−1
Then for x=
2
π
4cosy(
2
π
)=−e
−2
+1
Therefore,
198+e
−2
+4cosy(
2
π
)=198+e
−2
−e
−2
+1=199