If x- 0 y dt - 1-9 t 2 and d 2 y d x 2 -ay- then find a-

x=∫^y_0dy√(1+9t^2)⇒ differentiate ⇒ dx=dy√(1+9y^2)⇒dydx=√(1+9y^2) ⇒d^2ydx^2=9× 2y2√(1+9y^2)dydx=9y√(1+9y^2)√(1+9y^2) ⇒d^2ydx^2=9y ...

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Byju's Answer

Standard XII

Mathematics

Chain Rule of Differentiation

If x=∫ 0 y ...

Question

If $x = \int_{0}^{y} \frac{d t}{\sqrt{1 + 9 t^{2}}}$ and $\frac{d^{2} y}{d x^{2}} = a y$ , then find $a$ .

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Solution

$x = \int_{0}^{y} \frac{d y}{\sqrt{1 + 9 t^{2}}} \Rightarrow$ differentiate $\Rightarrow$
$d x = \frac{d y}{\sqrt{1 + 9 y^{2}}} \Rightarrow \frac{d y}{d x} = \sqrt{1 + 9 y^{2}}$
$\Rightarrow \frac{d^{2} y}{d x^{2}} = \frac{9 \times 2 y}{2 \sqrt{1 + 9 y^{2}}} \frac{d y}{d x} = \frac{9 y}{\sqrt{1 + 9 y^{2}}} \sqrt{1 + 9 y^{2}}$
$\Rightarrow \frac{d^{2} y}{d x^{2}} = 9 y$
$∴ a = 9$

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If x=∫y0dt√1+9t2 and d2ydx2=ay, then find a.

x=∫y0dy√1+9t2⇒ differentiate ⇒dx=dy√1+9y2⇒dydx=√1+9y2⇒d2ydx2=9×2y2√1+9y2dydx=9y√1+9y2√1+9y2⇒d2ydx2=9y∴a=9

If $x = \int_{0}^{y} \frac{d t}{\sqrt{1 + 9 t^{2}}}$ and $\frac{d^{2} y}{d x^{2}} = a y$ , then find $a$ .