Question 459 of 490MediumMCQ
Let y=y(x) be the solution of the differential equation
(3y2−5x2)y dx+2x(x2−y2)dy=0
such that y(1)=1. Then (y(2))3−12y(2) is equal to : Solutions · 1
Official solution
(3y2−5x2)y⋅dx+2x(x2−y2)dy=0
\[
\Rightarrow \frac{d y}{d x}=\frac{y\left(5 x^{2}-3 y^{2}\right)}{2 x\left(x^{2}-y^{2}\right)}
\]
Put y=mx
\[
\Rightarrow m+x \cdot \frac{d m}{d x}=\frac{m\left(5-3 m^{2}\right)}{2\left(1-m^{2}\right)}
\]
\[
x⋅dxdm=2(1−m2)(5−3m2)m−2m(1−m2)⇒xdx=m(m2−3)2( m2−1)dm⇒xdx=(m2−m34+ m2−334m)dm
\]
⇒∫xdx=∫m(32)+∫32(m2−32m)dm
⇒ln∣x∣=32ln∣m∣+32lnm2−3+C
Or, ln∣x∣=32lnxy+32ln(xy)2−3+C
Put (x=1,y=1) : we get c=−32ln(2)
⇒ln∣x∣=32lnxy+32ln(xy)2−3−32ln(2)
⇒(xy)[(xy)2−3]=2.(x3/2) ........(1)
Put x=2 in equation (1), we get
⇒y(y2−12)=4×2×2×22
⇒y3−12y=322
⇒y3(2)−12y(2)=322 Answer the question to unlock