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Continuity and Differentiability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity and Differentiability4 Marks
Question
If $y^3+3 a x^2+x^3=0$ then prove that:$
\frac{d^2 y}{d x^2}+\frac{2 a^2 x^2}{y^5}=0
$
✓

Answer

Given that :
$
y^3+3 a x^2+x^3=0
....(1)$
differentiating both sides w.r.t. $x$
$
3 y^2 \frac{d y}{d x}+6 a x+3 x^2=0
$
or $\quad$$
y^2 \frac{d y}{d x}+x^2+2 a x=0
......(2)$
or$\quad$$
\frac{d y}{d x}=-\left(\frac{x^2+2 a x}{y^2}\right) .
$
again differentiating both sides of (2) w.r.t. $x$$
\begin{aligned}
y^2 \frac{d^2 y}{d x^2}+2 y\left(\frac{d y}{d x}\right)^2 & =-2 x-2 a \\
y^2 \frac{d^2 y}{d x^2}+2 y\left(\frac{d y}{d x}\right)^2+2 x+2 a & =0 \\
y^2 \frac{d^2 y}{d x^2}+2\left[y\left(\frac{d y}{d x}\right)^2+x+a\right] & =0
\end{aligned}
$
from equation (3) put the value of $\frac{d y}{d x}$ :
or $y^2 \frac{d^2 y}{d x^2}+2\left[y\left(-\left(\frac{x^2+2 a x}{y^2}\right)\right)^2+x+a\right]=0$
or $y^2 \frac{d^2 y}{d x^2}+\frac{2}{y^3}\left[x^4+4 a^2 x^2+4 a x^3+(a+x) y^3\right]=0$
or $y^2 \frac{d^2 y}{d x^2}+\frac{2}{y^3}\left[x^4+4 a^2 x^2+4 a x^3+(a+x)\right.$
$\left.\left(-3 a x^2-x^3\right)\right]=0[$ from equation (1)]
or $y^2 \frac{d^2 y}{d x^2}+\frac{2}{y^3}\left[x^4+4 a^2 x^2+4 a x^3-3 a^2 x^2-3 a x^3\right.$
$\left.-a x^3-x^4\right]=0$
$y^2 \frac{d^2 y}{d x^2}+\frac{2}{y^3}\left(a^2 x^2\right)=0$
or $\quad \frac{d^2 y}{d x^2}+\frac{2 a^2 x^2}{y^5}=0 \quad$
Hence proved.

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