Question

If $y = \cos ecx + \cot x$, then find: $\sin x\dfrac{{{d^2}y}}{{d{x^2}}} + {y^2}$.

Answer 1

In this problem, first we need to find the first derivative of the given function. Then, find the second derivative of the function and simplify it by back substitution. Now substitute the obtained expressions into the given differential equation.

Complete step by step answer:
The first derivative of the function $y = \cos ecx + \cot x$ is obtained as shown below.

$$\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {\cos ecx + \cot x} \right) \\\ = \dfrac{d}{{dx}}\left( {\cos ecx} \right) + \dfrac{d}{{dx}}\left( {\cot x} \right) \\\ = \left( { - \cos ecx\cot x} \right) + \left( { - \cos e{c^2}x} \right) \\\ = - \cos ecx\left( {\cot x + \cos ecx} \right) \\\ = - y\cos ecx \\\$$

Now, find the second derivative of the function $y = \cos ecx + \cot x$ using the product rule as shown below.

$$\dfrac{d}{{dx}}\left( {\dfrac{{dy}}{{dx}}} \right) = - \dfrac{d}{{dx}}\left( {y\cos ecx} \right) \\\ \dfrac{{{d^2}y}}{{d{x^2}}} = - \left[ {y\dfrac{d}{{dx}}\left( {\cos ecx} \right) + \cos ecx\dfrac{d}{{dx}}\left( y \right)} \right] \\\ = - \left[ {y\left( { - \cos ecx\cot x} \right) + \cos ecx\dfrac{{dy}}{{dx}}} \right] \\\ = - \left[ { - y\cos ecx\cot x + \cos ecx\dfrac{{dy}}{{dx}}} \right] \\\$$

Now, substitute $- y\cos ecx$ for $\dfrac{{dy}}{{dx}}$ in the above expression.

$$\dfrac{{{d^2}y}}{{d{x^2}}} = - \left[ { - y\cos ecx\cot x + \cos ecx\left( { - y\cos ecx} \right)} \right] \\\ = - \left( { - y\cos ecx} \right)\left[ {\cot x + \cos ecx} \right] \\\ = y\cos ecx\left[ y \right] \\\ = {y^2}\cos ecx \\\$$

Now, substitute ${y^2}\cos ecx$ for $\dfrac{{{d^2}y}}{{d{x^2}}}$ in expression $\sin x\dfrac{{{d^2}y}}{{d{x^2}}} + {y^2}$.

$$\,\,\,\,\,\sin x\left( {{y^2}\cos ecx} \right) + {y^2} \\\ \Rightarrow \sin x\left( {{y^2} \cdot \dfrac{1}{{\sin x}}} \right) + {y^2} \\\ \Rightarrow {y^2} + {y^2} \\\ \Rightarrow 2{y^2} \\\$$

Thus, the expression for the differential equation $\sin x\dfrac{{{d^2}y}}{{d{x^2}}} + {y^2}$ is $2{y^2}$.

Note: Always use back substitution while obtaining the first and second derivatives in order to minimize the complexity. The second derivative of the function f is derivative of derivative of the function f. The second derivative measures the instantaneous rate of change of the first derivative. In other words the second derivative tells whether the slope of the tangent line is increasing or decreasing.