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 . &nbsp 2 y e 2 x = C . e 2 x + 1 B . &nbsp 2 y e 2 x = C . e 2 x − 1 C . &nbsp y e 2 x = C . e 2 x + 2 D . &nbsp 2 x e 2 y = C . e x − 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&#x

Discussion Forum : Differential Equations

Question -

Solution to the differential equation $\frac{x + \frac{x^{3}}{3!} + \frac{x^{5}}{5!} + . . . . .}{1 + \frac{x^{2}}{2!} + \frac{x^{4}}{4!} + . . . . .} = \frac{d x - d y}{d x + d y}$ is

Options:

A .

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

B .

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

C .

y e^{2 x} = C . e^{2 x} + 2

D .

2 x e^{2 y} = C . e^{x} - 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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Discussion Forum : Differential Equations

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