Let y prime=(4 y2+4 x y+x2/4 x2) and y(1)=0, then y(e(π/2)) equals

Question

Let y prime=(4 y2+4 x y+x2/4 x2) and y(1)=0, then y(e(π/2)) equals (A) (1/2) e(π/2) (B) e (π/2) (C) (1/4) e (π/2) (D) (π/2)

Tardigrade Admin · Accepted Answer

Put y=v x v+(x d v/d x)=(4 v2+4 v+1/4) ⇒ (x d v/d x)=(4 v2+1/4) ∫ (d v/v2+(1)4)= ln x+C 2 tan -1(2 v)= ln x+C x=1, y=0 ∴ C=0 2 v= tan ((1/2) ln x) y=(x/2) tan ((1/2) ln x) y(eπ / 2)=(eπ / 2/2)

Q. Let $y^{'} = \frac{4 y ^{2} + 4 x y + x ^{2}}{4 x ^{2}}$ and $y (1) = 0$ , then $y (e^{\frac{π}{2}})$ equals

$\frac{1}{2} e^{\frac{π}{2}}$

$e^{\frac{π}{2}}$

$\frac{1}{4} e^{\frac{π}{2}}$

$\frac{π}{2}$

Q. Let y′=4x24y2+4xy+x2​ and y(1)=0, then y(e2π​) equals

21​e2π​

e2π​

41​e2π​

2π​

Put y=vx v+dxxdv​=44v2+4v+1​⇒dxxdv​=44v2+1​ ∫v2+41​dv​=lnx+C 2tan−1(2v)=lnx+C x=1,y=0 ∴C=0 2v=tan(21​lnx) y=2x​tan(21​lnx) y(eπ/2)=2eπ/2​

Q. Let $y^{'} = \frac{4 y ^{2} + 4 x y + x ^{2}}{4 x ^{2}}$ and $y (1) = 0$ , then $y (e^{\frac{π}{2}})$ equals

$\frac{1}{2} e^{\frac{π}{2}}$

$e^{\frac{π}{2}}$

$\frac{1}{4} e^{\frac{π}{2}}$

$\frac{π}{2}$