Question

If $x = \int\limits_0^y {\dfrac{1}{{\sqrt {1 + 9{t^2}} }}} dt$ and $\dfrac{{{d^2}y}}{{d{x^2}}} = ay$ then $a$ is equal to…?

Answer 1

In order to achieve our solution or to find the value of $a$, here first we have to integrate the differentiation, which is nothing but using Leibniz Integral Rule is given below, $\dfrac{d}{{dx}}\left[ {\int\limits_{u(x)}^{v(x)} {f(x)} } \right] = \int {v(x)\dfrac{d}{{dx}}(v(x)) - } \int {u(x)\dfrac{d}{{dx}}(u(x))}$ and after that we have to differentiate with respect to y, which is partially completed. After that, we have to work on a second-degree derivative and by using chain rule to find the value of $a$.

Complete step by step answer:
We are given the integral function,

\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = ay \\\ $$ ---------(1) Here by using Leibniz Integral Rule, which says that Differentiation under the integral, which means this rule can be used to evaluate certain unusual definite integrals. So, by using the same rule, we have to find the solution to the given equation. $ \Rightarrow \dfrac{d}{{dx}}\left[ {\int\limits_{u(x)}^{v(x)} {f(x)} } \right] = \int {v(x)\dfrac{d}{{dx}}(v(x)) - } \int {u(x)\dfrac{d}{{dx}}(u(x))} $ Now, we have to differentiate the equation (1) with respect to y, we get: $$ \Rightarrow \dfrac{{dx}}{{dy}} = \dfrac{1}{{\sqrt {1 + 9{y^2}} }}.\dfrac{{dy}}{{dy}} - \dfrac{1}{{\sqrt {1 + 9{{(0)}^2}} }}\dfrac{d}{{dy}}(0)$$ Now, we have to reciprocate the $\dfrac{{dx}}{{dy}}$, then we will get: $ \Rightarrow \dfrac{{dy}}{{dx}} = \sqrt {1 + 9{y^2}} $ Now, we have got $\dfrac{{dx}}{{dy}}$, but actually we have to calculate for $\dfrac{{{d^2}y}}{{d{x^2}}}$, which will be as follows: Now, again we have to differentiate with respect to x, we will get: $ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{d}{{dx}}{(1 + 9{y^2})^{\dfrac{1}{2}}}$. Since, we can write $\sqrt {1 + 9{y^2}} = {(1 + 9{y^2})^{\dfrac{1}{2}}}$. $$ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{1}{2}.{(1 + 9{y^2})^{\dfrac{1}{2} - \dfrac{1}{2}}}18y$$ since, $\sqrt {1 + 9{y^2}} = {(1 + 9{y^2})^{\dfrac{1}{2}}},{(1 + 9{y^2})^{ - \dfrac{1}{2}}} = \dfrac{1}{{\sqrt {1 + 9{y^2}} }}$. We have got the above equation using the chain rule, which tells us how to differentiate composite functions. $$ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{9y\sqrt {1 + 9{y^2}} }}{{\sqrt {1 + 9{y^2}} }}$$ $$ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = 9y$$ Now, from the above equation, we can say that: $$ \Rightarrow 9y = ay$$ $$ \Rightarrow 9 = a$$ $$ \therefore a = 9$$ **Hence, if $$ x = \int\limits_0^y {\dfrac{1}{{\sqrt {1 + 9{t^2}} }}} dt $$ and $$\dfrac{{{d^2}y}}{{d{x^2}}} = ay$$ then a is equal to $$9$$.** **Note:** We note that, to differentiate a composite function, which is the only chain rule we have. If we do not know whether a function is composite or not, then we will not be able to differentiate correctly. So, from the above given question we have used the chain rule of Leibniz notation to achieve our solution.