plz sol... my question
(1-x*x) dy/dx +xy = xy*y how to solve this??
Answers
Answer:
(1+x)
dx
dy
−xy=1−x
dx
dy
+(
1+x
−x
)y=
1+x
1−x
∫pdx=∫
1+x
−x
=−∫
1+x
1+x−1
dx
=−∫1−
1+x
1
dx
=−x+log∣1+x∣dx
I.Fe
∫pdx
=e
log(1+x)−x
=
e
x
1+x
4.5y.
e
x
1+x
=∫
e
x
1−x
y
e
x
(1+x)
=∫e
−x
(1−x)
y(1+x)=e
x
c
−x
(+x)
y(1+x)=x
Step-by-step explanation:
Answer:
If the question is like this :-
Solution (1) :-
dx/dy=(1-x)/y
or 1/y.dy = 1/(1-x).dx
logy = -log (1-x) +logC . = log C/(1-x)
or y = C/(1-x)
y.(1-x) = C , Answer.
If the question is like this :-
Solution (2nd) :-
dx/dy = 1 - ( x/y) = (y-x)/y
or dy/dx= y/(y-x)…………(1)
Let y = vx
dy/dx= v.1+x.dv/dx…………..(2)
On putting dy/dx = v+x.dv/dx from eq.(2)
v+x.dv/dx= vx/(vx-x) = v/(v-1)
x.dv/dx = v/(v-1) -v =(v-v^2+v)/(v-1)
-(v-1)/(v^2-2v).dv = 1/x.dx.
(1/2).[(2v-2)/(v^2–2v).dv] = -1/x.dx
(1/2).log (v^2–2v) = -log x +log C = log C/x
log (v^2–2v)^1/2 = log C/x
ot (v^2–2v)^1/2 = C/x
On putting v= y/x
or ( y^2/x^2 - 2y/x)^1/2 = C/x
or [(y^2–2xy)^1/2]/x = C/x
or (y^2–2xy)^1/2 = C.
or y^2–2xy = C^2
or y^2 -2xy = k . Answer