Find the equation of a curve passing through the point (0, 0) and whose differential equation is
e sin x.
dx
dy
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Answer:
ANSWER
given,
dx
dy
=e
x
sinx
dy=e
x
sinx.dx
∫dy=∫e
x
sinx.dx → integrating both sides.
det, ∫e
x
sinx dx=I−−−−(1)
y=∫
f(x)
↓
e
x
g(x)
↓
sinx
dx=I
Since, ∫f(x)g(x)dx=f(x).∫g(x)dx−∫[f
′
(x)∫g(x)dx]dx
∴ ⇒ e
x
.∫sinx dx−∫[e
x
.∫sinx dx]
I=e
x
.(−cosx)−∫e
x
.(−cosx).dx
I=−e
x
cosx+∫
f(x)
↓
e
x
.
g(x)
↓
cosx
dx
Now apply same formula as above
I=−e
x
cosx.+[e
x
.∫cosx.dx−∫(ex.∫cosx.dx)dx]
I=−e
x
.cosx+
⎣
⎢
⎢
⎢
⎢
⎡
e
x
.sinx−
(I)
∫e
x
.sinx.dx
⎦
⎥
⎥
⎥
⎥
⎤
I=−e
x
cosx+e
x
sinx−I+c
2y=e
x
(sinx−cosx)+c
Since this curve passes through (0,0) therefore
2×0=e
0
(sin0cos0)+c
c=1
∴ equation of curve 2y=e
x
(sinx−cosx)+1
(2y−1)=e
x
(sinx−cosx)
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