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Question : 75 of 160
Marks:
+1,
-0
Solution:
x‌=y(log‌y−log‌x+1) ‌=‌(log‌+1)...(i)
Let
f(x,y)=‌(log‌+1) f(kx,ky)=‌(log‌+1) ∴ Given differentiate Equation is homogeneous
{| ∴put‌,‌y=vx |
| ‌=v+x‌ |
} From Eq. (i),
v+x‌=‌(log‌+1) v+x‌=v(log‌v+1) v+x‌=v‌log‌v+v ‌dv=‌dx Integrating on both sides,
∫‌⋅‌dv=∫‌dx Put,
‌‌log‌v=t ‌dv=dt ∫‌dt=∫‌dx log‌t=log‌x+log‌c log(log‌v)=log‌x‌c logev=xc v=exc ‌=exc y=x.exc
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