Solve: x ( 1 + y2) dx - y ( 1 + x2 ) dy = 0, given that y - Maths - Differential Equations

Question

Solve:
x ( 1 + y2) dx - y ( 1 + x2 ) dy = 0,
given that y = 0 when x = 1
The answer given in the book is : ( 1 + x2 ) = 2 ( 1
+ y2 )

Manbar Singh · Accepted Answer

The given differential equation is,
x1+ y2 dx - y1 + x2dy = 0⇒dydx = x1+ y2 y1 + x2⇒y1 + y2dy
= x1 + x2 dx [VARIABLE SEPARABLE]Integrating both sides we get,
∫y1 + y2dy = ∫x1 + x2 dx (1)Let I1 = ∫y1 + y2dyput 1 +
y2 = u ⇒2y dy = du ⇒y dy = 12du
so, I1 = 12∫duuI1 = logu + C1I1 = log1 + y2 + C1

Let I2 = ∫x1 + x2dxput 1 + x2 = v⇒2x dx = dv⇒x dx =
12dvso, I2 =12 ∫dvv= 12log (v) + C2 = 12log(1 +x2) + C2

from (1), we get
12log (1 + y2) + C1 = 12log(1 + x2) + C2⇒12log (1 + y2) =
12log(1 + x2) + C2 - C1⇒12log (1 + y2) = 12log(1 + x2) + C (2)
[where C2 - C1 = C]put x = 1 and y = 0 in (2), we get12log 1 =
12log 2 + C⇒0 = 12log 2 + C [as, log 1 = 0]So, C =-12 log
2

putting the value of C in (2), we get

12log(1 + y2) = 12log(1 + x2) - 12log 2⇒log(1 + y2) = log(1 +
x2) - log 2⇒log(1 + y2) = log1 + x22⇒1 + y2 = 1 +
x22⇒1 + x2 = 2(1 + y2)

Solve: x ( 1 + y2) dx - y ( 1 + x2 ) dy = 0, given that y = 0 when x = 1 The answer given in the book is : ( 1 + x2 ) = 2 ( 1 + y2 )

Solve:

x ( 1 + y²) dx - y ( 1 + x² ) dy = 0, given that y = 0 when x = 1

The answer given in the book is : ( 1 + x² ) = 2 ( 1 + y² )