Solutions of the differential equation dy-dx 2 - dy-dx ex - e-x - 1 - 0 are given by-

The correct options are A y + e^ x = C B y e^x = C ( dy dx #160;) #160;^ 2 dy dx ( e ^ x + e ^ x )+1=0 ⇒ ( dy dx e ^ x )( ...

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Standard XII

Mathematics

Linear Differential Equations of First Order

Solutions of ...

Question

Solutions of the differential equation ${(\frac{d y}{d x})}^{2} - \frac{d y}{d x} (e^{x} + e^{- x}) + 1 = 0$ are given by:

y + e^{- x} = C

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y - e^{- x} = C

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y + e^{x} = C

No worries! We‘ve got your back. Try BYJU‘S free classes today!

y - e^{x} = C

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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Solution

The correct options are
A $y + e^{- x} = C$
B $y - e^{x} = C$
${(\frac{d y}{d x})}^{2} - \frac{d y}{d x} (e^{x} + e^{- x}) + 1 = 0 \Rightarrow (\frac{d y}{d x} - e^{x}) (\frac{d y}{d x} - e^{- x}) = 0 \Rightarrow y = e^{x} + c a n d y = - e^{- x} + c$

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Similar questions

The general solution of the differential equation $\frac{d y}{d x} = e^{x + y}$ is
(a) $e^{x} + e^{- y} = C$
(b) $e^{x} + e^{y} = C$
(c) $e^{- x} + e^{y} = C$
(d) $e^{- x} + e^{- y} = C$

Q. The general solution of the differential equation

\frac{d y}{d x} = e^{x + y}

, is
(a) e^x+ e^−y = C
(b) e^x + e^y = C
(c) e⁻^x + e^y = C
(d) e^−x + e^−y = C

The general solution of the differential equation $e^{x} d y + (y e^{x} + 2 x) d x = 0$ is

a) $x e^{y} + x^{2} = C$

b) $x e^{y} + y^{2} = C$

c) $y e^{x} + x^{2} = C$

d) $y e^{y} + x^{2} = C$

Q. The solution of the differential equation

x \frac{d y}{d x} = y (l o g y - l o g x + 1)

Q. The general solution of the differential equation

(1 + e^{\frac{x}{y}}) d x + (1 - \frac{x}{y}) e^{\frac{x}{y}} d y = 0

is (

c

is an arbitary constant)

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Solutions of the differential equation (dydx)2−dydx(ex+e−x)+1=0 are given by:

The correct options are A y+e−x=C B y−ex=C(dydx)2−dydx(ex+e−x)+1=0⇒(dydx−ex)(dydx−e−x)=0⇒y=ex+candy=−e−x+c

Solutions of the differential equation ${(\frac{d y}{d x})}^{2} - \frac{d y}{d x} (e^{x} + e^{- x}) + 1 = 0$ are given by:

The correct options are
A $y + e^{- x} = C$
B $y - e^{x} = C$
${(\frac{d y}{d x})}^{2} - \frac{d y}{d x} (e^{x} + e^{- x}) + 1 = 0 \Rightarrow (\frac{d y}{d x} - e^{x}) (\frac{d y}{d x} - e^{- x}) = 0 \Rightarrow y = e^{x} + c a n d y = - e^{- x} + c$