The integrating factor of the differential equation x - y 2 dy - dx -1- A- e-yB- e-xC- exD- e y

The correct option is A e^ y(x + y^2)dy/dx = 1 ⇒dx/dy = x +y^2 ⇒ nbsp;dx/dy nbsp; x = y^2 Itegrating factor = e^∫ 1 dy = e^ y ...

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

Byju's Answer

Standard XII

Mathematics

Degree of a Differential Equations

The integrati...

Question

The integrating factor of the differential equation $(x + y^{2}) \frac{d y}{d x} = 1$ is

e^{- x}

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

e^{- y}

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

e^{x}

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

e^{y}

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

Open in App

Solution

The correct option is B $e^{- y}$
$(x + y^{2}) \frac{d y}{d x} = 1 \Rightarrow \frac{d x}{d y} = x + y^{2} \Rightarrow \frac{d x}{d y} - x = y^{2}$
Itegrating factor
$= e^{\int - 1 d y} = e^{- y}$

Suggest Corrections

Similar questions

The integrating factor (IF) of the differential equation
$(1 - y^{2}) \frac{d y}{d x} + y x = a y (- 1 < y < 1)$ is
(a) $\frac{1}{y^{2} - 1}$
(b) $\frac{1}{\sqrt{y^{2} - 1}}$
(c) $\frac{1}{1 - y^{2}}$
(d) $\frac{1}{\sqrt{1 - y^{2}}}$

Q. The integrating factor of the differential equation

(1 + x^{2}) \frac{d y}{d x} + y = e^{{tan}^{- 1} x}

is:

Q. The integrating factor of

\frac{d y}{d x} + y = \frac{1 + y}{x}

Q. The integrating factor of the differential equation

3 x {log}_{e} x \frac{d y}{d x} + y = 2 {log}_{e} x

is given by:

Q. The integrating factor of the differential equation

x \frac{d y}{d x} - y = 2 x^{2}

(a) e^−x

(b) e⁻^y

(c)

\frac{1}{x}

(d) x

Join BYJU'S Learning Program

The integrating factor of the differential equation (x+y2)dydx=1 is

The correct option is B e−y(x+y2)dydx=1⇒dxdy=x+y2⇒dxdy−x=y2 Itegrating factor =e∫−1 dy=e−y

The integrating factor of the differential equation $(x + y^{2}) \frac{d y}{d x} = 1$ is

The correct option is B $e^{- y}$
$(x + y^{2}) \frac{d y}{d x} = 1 \Rightarrow \frac{d x}{d y} = x + y^{2} \Rightarrow \frac{d x}{d y} - x = y^{2}$
Itegrating factor
$= e^{\int - 1 d y} = e^{- y}$