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Mathematics Question on Differential equations

For the curve
C:(x2+y2–3)+(x2–y2–1)5=0,\begin{array}{l}C : \left(x^2 + y^2 – 3\right) + \left(x^2 – y^2 – 1\right)^5 = 0, \end{array}

the value of 3y’–y3y”,at the point\begin{array}{l}\text{the value of}\ 3y’ – y^3y”, \text{at the point}\end{array}

(α,α),α>0(α, α), α > 0, on C is equal to ________.

Answer

∵C:(x2+y2–3)+(x2–y2–1)5=0\begin{array}{l}\because C : \left(x^2 + y^2 – 3\right) + \left(x^2 – y^2 – 1\right)^5 = 0\end{array}for point (α,α)(α, α).
α2+α2–3+(α2–α2–1)5=0\begin{array}{l}\alpha ^2 + \alpha ^2 – 3 + (\alpha^ 2 – \alpha ^2 – 1)^5 = 0\end{array}
∴ α=2\begin{array}{l} \therefore\ \alpha=\sqrt{2}\end{array}
On differentiating(x2+y2–3)+(x2–y2–1)5=0 we get x+yy’+5(x2–y2–1)4(x–yy’)=0…(i)\begin{array}{l}\text{On differentiating} \left(x^2 + y^2 – 3\right) + \left(x^2 – y^2 – 1\right)^5 = 0 ~\text{we get}\\\ x + yy’ + 5 \left(x^2 – y^2 – 1\right)^4 \left(x – yy’\right) = 0 …\left(i\right)\end{array}
Whenx=y=2 \begin{array}{l} x=y=\sqrt{2}\end{array}then y’=32\begin{array}{l} y’=\frac{3}{2} \end{array}
Again on differentiating eq. (i) we get :
1+(y’)2+yy”+20(x2–y2–1)(2x–2yy’)\begin{array}{l}1 + \left(y’\right)^2 + yy” + 20 \left(x^2 – y^2 – 1\right) \left(2x – 2 yy’\right) \end{array}
(x–y’y)+5(x2–y2–1)4(1–y’2–yy”)=0\begin{array}{l}\left(x – y’y\right) + 5\left(x^2 – y^2 – 1\right)^4 \left(1 – y’^2 – yy”\right) = 0\end{array}
Forx=y=2 \begin{array}{l} x=y=\sqrt{2} \end{array}and y’=32\begin{array}{l} y’=\frac{3}{2}\end{array}
we get y”=−2342\begin{array}{l} y”=-\frac{23}{4\sqrt{2}}\end{array}
\begin{array}{l}\therefore 3y’ – y^3y” =3\cdot\frac{3}{2}-\left(\sqrt{2}\right)^3\cdot\left(-\frac{23}{4\sqrt{2}}\right)\\\= 16\end{array}