Question

The solution of the differential equation \[{x^2}\dfrac{{dy}}{{dx}}\cos \left( {\dfrac{1}{x}} \right) - y\sin \left( {\dfrac{1}{x}} \right) = - 1\]; where \[y \to - 1\] as \[x \to \infty \] is
A) \[y = \sin \left( {\dfrac{1}{x}} \right) - \cos \left( {\dfrac{1}{x}} \right)\]
B) \[y = \dfrac{{x + 1}}{{x\sin \left( {\dfrac{1}{x}} \right)}}\]
C) \[y = \cos \left( {\dfrac{1}{x}} \right) + \sin \left( {\dfrac{1}{x}} \right)\]
D) \[y = \dfrac{{x + 1}}{{x\cos \left( {\dfrac{1}{x}} \right)}}\]

Answer 1

Hint:
Here we need to solve the given differential equation. We will first rearrange the terms and after rearranging the terms, we will find that the given differential equation is a linear differential equation. Then we will find the integrating factor, at last we will find the solution of this linear differential equation using the formula.

Complete step by step solution:
Here we need to solve the given differential equation and the given differential equation is:-
\[{x^2}\dfrac{{dy}}{{dx}}\cos \left( {\dfrac{1}{x}} \right) - y\sin \left( {\dfrac{1}{x}} \right) = - 1\]
Now, we will first rearrange the terms.
\[ \Rightarrow {x^2}\dfrac{{dy}}{{dx}}\cos \left( {\dfrac{1}{x}} \right) = - 1 + y\sin \left( {\dfrac{1}{x}} \right)\]
Now, we will divide both sides by the term \[{x^2}\cos \left( {\dfrac{1}{x}} \right)\].
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - 1}}{{{x^2}\cos \left( {\dfrac{1}{x}} \right)}} + \dfrac{{y\sin \left( {\dfrac{1}{x}} \right)}}{{{x^2}\cos \left( {\dfrac{1}{x}} \right)}}\]
We know from the inverse trigonometric identities that
\[\begin{array}{l}\sec \theta = \dfrac{1}{{\cos \theta }}\\\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\end{array}\]
Using these trigonometric identities, we get
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - \sec \left( {\dfrac{1}{x}} \right)}}{{{x^2}}} + \dfrac{{y\tan \left( {\dfrac{1}{x}} \right)}}{{{x^2}}}\]
On further simplification, we get
\[ \Rightarrow \dfrac{{dy}}{{dx}} - \dfrac{{\tan \left( {\dfrac{1}{x}} \right)}}{{{x^2}}}y = \dfrac{{ - \sec \left( {\dfrac{1}{x}} \right)}}{{{x^2}}}\]
We can see that the differential equation is in the form \[\dfrac{{dy}}{{dx}} + Py = Q\] …………… \[\left( 1 \right)\]
And the solution of such a type of differential equation is\[y \times I.F = \int {Q \times I.Fdy} + c\].
Where, \[I.F\left( {{\rm{Integrating factor}}} \right) = {e^{\int {Pdx} }}\]
We will follow the same step to solve this differential equation i.e.
\[ \Rightarrow \dfrac{{dy}}{{dx}} - \dfrac{{\tan \left( {\dfrac{1}{x}} \right)}}{{{x^2}}}y = \dfrac{{ - \sec \left( {\dfrac{1}{x}} \right)}}{{{x^2}}}\]
On comparing this differential equation with the differential equation in equation 1, we get
\[\begin{array}{l}P = - \dfrac{{\tan \left( {\dfrac{1}{x}} \right)}}{{{x^2}}}\\Q = \dfrac{{ - \sec \left( {\dfrac{1}{x}} \right)}}{{{x^2}}}\end{array}\]
We will first find the integrating factor.
\[I.F = {e^{\int {Pdx} }} = {e^{\int {\dfrac{{ - \tan \left( {\dfrac{1}{x}} \right)}}{{{x^2}}}dx} }}\]
Let \[\dfrac{1}{x} = t\]
On differentiating the terms, we get
\[ \Rightarrow - \dfrac{1}{{{x^2}}}dx = dt\]
On substituting this value in the exponent, we get
\[I.F = {e^{\int {\tan t \cdot dt} }}\]
On integrating the terms, we get
\[ \Rightarrow I.F = {e^{\ln \left( {\sec t} \right)}} = {e^{\ln \left( {\sec \left( {\dfrac{1}{x}} \right)} \right)}}\]
We know from the properties of exponentials that \[{e^{\ln x}} = x\]
Using this property here, we get
\[ \Rightarrow I.F = \sec \left( {\dfrac{1}{x}} \right)\]
Therefore, the solution of this differential equation is equal to \[y \times I.F = \int {Q \times I.Fdy} + c\]
On substituting the value of the integrating factor and the constant values we get
\[y \times \sec \left( {\dfrac{1}{x}} \right) = \int {\dfrac{{ - \sec \left( {\dfrac{1}{x}} \right)}}{{{x^2}}} \times \sec \left( {\dfrac{1}{x}} \right)dx} + c\]
On further simplifying the terms, we get
\[ \Rightarrow y \times \sec \left( {\dfrac{1}{x}} \right) = \int {\dfrac{{ - {{\sec }^2}\left( {\dfrac{1}{x}} \right)}}{{{x^2}}}dx} + c\]
Let \[\dfrac{1}{x} = t\]
On differentiating the terms, we get
\[ \Rightarrow - \dfrac{1}{{{x^2}}}dx = dt\]
Now, we will substitute this value in the integration, we get
\[ \Rightarrow y \times \sec \left( {\dfrac{1}{x}} \right) = \int {{{\sec }^2}t \cdot dt} + c\]
On integrating the terms, we get
\[ \Rightarrow y \times \sec \left( {\dfrac{1}{x}} \right) = \tan t + c\]
Now, we will substitute the value of \[t\] here.
\[ \Rightarrow y \times \sec \left( {\dfrac{1}{x}} \right) = \tan \left( {\dfrac{1}{x}} \right) + c\] ……………….. \[\left( 2 \right)\]
It is given that \[y \to - 1\] and \[x \to \infty \]
We can say that \[\dfrac{1}{x} \to 0\]
Now, we will substitute this value in equation 2.
\[ \Rightarrow \left( { - 1} \right) \times \sec \left( 0 \right) = \tan \left( 0 \right) + c\]
We know that the value of
\[\begin{array}{l}\sec \left( 0 \right) = 1\\\tan \left( 0 \right) = 0\end{array}\]
Now, we will substitute these values here.
\[\begin{array}{l} \Rightarrow \left( { - 1} \right) \times 1 = 0 + c\\ \Rightarrow - 1 = c\end{array}\]
Now, we will substitute this value in equation 2.
\[\begin{array}{l} \Rightarrow y \times \sec \left( {\dfrac{1}{x}} \right) = \tan \left( {\dfrac{1}{x}} \right) + \left( { - 1} \right)\\ \Rightarrow y \times \sec \left( {\dfrac{1}{x}} \right) = \tan \left( {\dfrac{1}{x}} \right) - 1\end{array}\]
Now, we will divide both sides by the term \[\sec \left( {\dfrac{1}{x}} \right)\].
\[ \Rightarrow \dfrac{{y \times \sec \left( {\dfrac{1}{x}} \right)}}{{\sec \left( {\dfrac{1}{x}} \right)}} = \dfrac{{\tan \left( {\dfrac{1}{x}} \right)}}{{\sec \left( {\dfrac{1}{x}} \right)}} - \dfrac{1}{{\sec \left( {\dfrac{1}{x}} \right)}}\]
We know from the trigonometric formulas that:
\[\begin{array}{l}\dfrac{{\tan \theta }}{{\sec \theta }} = \sin \theta \\\cos \theta = \dfrac{1}{{\sec \theta }}\end{array}\]
Using these formulas here, we get
\[ \Rightarrow y = \sin \left( {\dfrac{1}{x}} \right) - \cos \left( {\dfrac{1}{x}} \right)\]

Hence, the correct option is option A.

Note:
A differential equation is defined as the equation which contains one or more than one functions like trigonometric functions, logarithmic functions, exponential functions etc with their derivatives. It is an inverse function of integration. The derivative of a function is defined as the rate of change of a function at a point. Derivation and its application is used in many fields like engineering, biology etc.