Solution to the differential equation x + x 3 3 ! + x 5 5 ! + . . . . . 1 + x 2 2 ! + x 4 4 ! + . . . . . = d x − d y d x + d y is Options: A . &nbsp2ye2x=C.e2x+1 B . &nbsp2ye2x=C.e2x−1 C . &nbspye2x=C.e2x+2 D . &nbsp2xe2y=C.ex−1

Answer: Option B : B Applying C and D, we get d y d x = e − x e x = e − 2 x ⇒ 2 y = − e − 2 x + C or 2 y e 2 x = C . e 2 x − 1 .

Solution to the differential equation x+x33!+x55!+.....1+x22!+x44!+.....=dx"

Question

Solution to the differential equation

\frac{x + \frac{x^{3}}{3!} + \frac{x^{5}}{5!} + . . . . .}{1 + \frac{x^{2}}{2!} + \frac{x^{4}}{4!} + . . . . .} = \frac{d x - d y}{d x + d y}

Options:

A . 2ye2x=C.e2x+1

B . 2ye2x=C.e2x−1

C . ye2x=C.e2x+2

D . 2xe2y=C.ex−1

Answer: Option B
:
B
Applying C and D, we get

\frac{d y}{d x} = \frac{e^{- x}}{e^{x}} = e^{- 2 x} \Rightarrow 2 y = - e^{- 2 x} + C

2 y e^{2 x} = C . e^{2 x} - 1

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