solve this: x(1+y^2)dx+y(1+x^2)dy=0
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The given differential equation is,
x(1+ y2) dx − y(1 + x2)dy = 0⇒dydx = x(1+ y2) y(1 + 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 = log∣∣u∣∣ + C1I1 = log∣∣1 + 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) = log(1 + x22)⇒1 + y2 = 1 + x22⇒1 + x2 = 2(1 + y2)
x(1+ y2) dx − y(1 + x2)dy = 0⇒dydx = x(1+ y2) y(1 + 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 = log∣∣u∣∣ + C1I1 = log∣∣1 + 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) = log(1 + x22)⇒1 + y2 = 1 + x22⇒1 + x2 = 2(1 + y2)
hemonshri73:
thanks for your help
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