Question

Solve the equation
X
(1+y²) dx+y(1 + x^2
) dy=
0​

Answer 1

According to the conditions,

= \frac{ x + 17}{x + 8 - 1} = \frac{3}{2}=

x+8−1

x+17

=

2

3

= > \frac{x + 17}{x + 7} = \frac{3}{2}=>

x+7

x+17

=

2

3

\begin{gathered} = > \frac{(x + 17)}{(x + 7)} \times (x + 7) = \frac{3}{2} \times (x + 7) \\ \\ (Multiplying \: both \: sides \: by \: (x + 7).)\end{gathered}

=>

(x+7)

(x+17)

×(x+7)=

2

3

×(x+7)

(Multiplyingbothsidesby(x+7).)

= > x + 17 = \frac{3}{2} (x + 7)=>x+17=

2

3

(x+7)

\begin{gathered} = > 2x(x + 17) = 2 \times \frac{3}{2} (x + 7) \\ \\ \: \: \: \: \: \: \: \: (Multiplying \: both \: sides \: by \: 2.)\end{gathered}

=>2x(x+17)=2×

2

3

(x+7)

(Multiplyingbothsidesby2.)

= > 2x + 34 = 3x + 21=>2x+34=3x+21

\begin{gathered} = > 2x - 3x = 21 - 34 \\ \\ (Transporting \: 3x \: to \: L.H.S. \: and \: 34 \: to \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: R.H.S.)\end{gathered}

=>2x−3x=21−34