Exercise 10.4-Additional Questions - Chapter 10 - Differentiability and Methods of Differentiation - 11th Maths Guide Samacheer Kalvi Solutions

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Additional Problems
Question 1.
If $y=A \cos 4 x+B \sin 4 x, A$ and $B$ are constants then Show that $y_2+16 y=0$
Solution:
$
\begin{aligned}
y_1 & =\frac{d y}{d x}=-4 \mathrm{~A} \sin 4 x+4 \mathrm{~B} \cos 4 x \\
y_2 & =\frac{d^2 y}{d x^2}=\frac{d}{d x}\left(\frac{d y}{d x}\right) \\
& =\frac{d}{d x}(-4 \mathrm{~A} \sin 4 x+4 \mathrm{~B} \cos 4 x) \\
& =-16 \mathrm{~A} \cos 4 x-16 \mathrm{~B} \sin 4 x \\
& =-16(\mathrm{~A} \cos 4 x+\mathrm{B} \sin 4 x)=-16 y \\
\therefore y_2+16 y & =0
\end{aligned}
$

Question 2.
If $y=\cos \left(m \sin ^{-1} x\right)$, prove that $\left(1-x^2\right) y_3-3 y_2+\left(m^2-1\right) y_1=0$
Solution:
We have $y=\cos \left(m \sin ^{-1} x\right)$
$\mathrm{y}_1=\sin \left(\mathrm{m} \sin ^{-1} \mathrm{x}\right) \cdot \frac{m}{\sqrt{1-x^2}}$
$
y_1^2=\sin ^2\left(m \sin ^{-1} x\right) \frac{m^2}{\left(1-x^2\right)}
$
This implies $\left(1-x^2\right) y_1^2=m^2 \sin ^2\left(m \sin ^{-1} x\right)=m^2\left[1-\cos ^2\left(m \sin ^{-1} x\right)\right]$
This is, $\left(1-x^2\right) y_1^2=m^2\left(1-y^2\right)$.
Again differentiating,
$
\begin{aligned}
\left(1-x^2\right) 2 y_1 \frac{d y_1}{d x}+y_1^2(-2 x) & =m^2\left(-2 y \frac{d y}{d x}\right) \\
\left(1-x^2\right) 2 y_1 y_2-2 x y_1^2 & =-2 m^2 y y_1 \\
\left(1-x^2\right) y_2-x y_1 & =m^2 y
\end{aligned}
$
Once again differentiating,
$
\begin{aligned}
& \left(1-x^2\right) \frac{d y_2}{d x}+y_2(-2 x)-\left[x \cdot \frac{d y_1}{d x}+y_1 \cdot 1\right]=-m^2 \frac{d y}{d x} \\
& \left(1-x^2\right) y_3-2 x y_2-x y_2-y_1=-m^2 y_1 \\
& \left(1-x^2\right) y_3-3 x y_2+\left(m^2-1\right) y_1=0
\end{aligned}
$