Solution to the differential equation x-x3-3-x5-5-1-x2-2-x4-4-dx-dy-dx-dy is

Applying C and D, we get dy/dx=e^ x/e^x=e^ 2x⇒ 2y= e^ 2x+C or 2y e^2x=C.e^2x 1. ...

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Standard XIII

Mathematics

Variable Separable Method

Solution to t...

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

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

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

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

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

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Solution

The correct option is B $2 y e^{2 x} = C . e^{2 x} - 1$

Applying C and D, we get
$\frac{d y}{d x} = \frac{e^{- x}}{e^{x}} = e^{- 2 x} \Rightarrow 2 y = - e^{- 2 x} + C$
or $2 y e^{2 x} = C . e^{2 x} - 1$ .

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Similar questions

Q. 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}

Q. 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}

Q. 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}

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

Q. If

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

then

\frac{d y}{d x} =

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Solution to the differential equation x+x33!+x55!+.....1+x22!+x44!+.....=dx−dydx+dy is

The correct option is B 2ye2x=C.e2x−1 Applying C and D, we get dydx=e−xex=e−2x⇒2y=−e−2x+C or 2ye2x=C.e2x−1.

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

The correct option is B $2 y e^{2 x} = C . e^{2 x} - 1$

Applying C and D, we get
$\frac{d y}{d x} = \frac{e^{- x}}{e^{x}} = e^{- 2 x} \Rightarrow 2 y = - e^{- 2 x} + C$
or $2 y e^{2 x} = C . e^{2 x} - 1$ .