What is the solution of the equation lndy-dx-x-0- where c is an arbitrary constant-

The correct option is C y+e^ x=c Given Differential equation is ln( dy dx #160; ) #160; +x=0 ⟹ dy dx = e ^ x Separating the variables ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Solving Linear Differential Equations of First Order

What is the s...

Question

What is the solution of the equation $l n (\frac{d y}{d x}) + x = 0$ ? where c is an arbitrary constant.

y + e^{x} = c

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

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

y - e^{x} = c

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

Open in App

Solution

The correct option is C $y + e^{- x} = c$
Given Differential equation is $ln (\frac{d y}{d x}) + x = 0 ⟹ \frac{d y}{d x} = e^{- x}$
Separating the variables and integrating them gives,
$\int e^{- x} d x = \int d y$
$⟹ - e^{- x} = y + c$ Where $c$ is the arbitrary constant
$⟹ y + e^{- x} = c$ is the general solution and $c$ is the arbitrary constant.

Suggest Corrections

Similar questions

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

(x^{2} y + e^{x}) d x - e^{x} d y = 0

is ___ {c is arbitrary constant}

Let $c$ be the arbitrary constant, then the solution of the differential equation $e^{x} cot y d x + (1 - e^{x}) {cosec}^{2} y d y = 0$ is

Q. The solution(s) of the differential equation

{(\frac{d y}{d x})}^{2} + 2 y cot x (\frac{d y}{d x}) = y^{2}

is/are

Join BYJU'S Learning Program

What is the solution of the equation ln(dydx)+x=0? where c is an arbitrary constant.

The correct option is C y+e−x=cGiven Differential equation is ln(dydx)+x=0⟹dydx=e−xSeparating the variables and integrating them gives,∫e−xdx=∫dy⟹−e−x=y+c Where c is the arbitrary constant⟹y+e−x=c is the general solution and c is the arbitrary constant.

What is the solution of the equation $l n (\frac{d y}{d x}) + x = 0$ ? where c is an arbitrary constant.