Question

Find the particular solution of the differential equation, \[\dfrac{{dy}}{{dx}} + y\cot x = 4x\cos ecx,x \ne 0\] given that \[y = 0\] when \[x = \dfrac{\pi }{2}\].

Answer 1

Hints: To solve this question, we mostly use the technique of an integrating factor. In this technique first, we have to determine the integrating factor from the differential equation. Then we must multiply both the sides of the differential equation by the integrating factor. By applying the product rule of differentiation and then integrating both sides the solution of the differential equation can be determined. As we have to obtain the particular solution of the differential equation we must have the value of integration constant which can be obtained by substituting the respective values of x and y given in the question in the solution.

Complete step-by-step solution:
The differential equation to be solved is given by,
\[\dfrac{{dy}}{{dx}} + y\cot x = 4x\cos ecx\]……..…………………………… (1)
Consider the general form of differential equation given by,
\[\dfrac{{dy}}{{dx}} + P(x)y = Q(x)\] ………………………………… (2)
Comparing eq. (1) and (2) we’ll get,
\[P(x) = \cot x\] ………………………………… (3)
And \[Q(x) = 4x\cos ecx\] ………………………………… (4)
Now we will find out the integrating factor which can be obtained by the formula given by
\[I = {e^{\int {P(x)dx} }}\] ………………………………… (5)
Substituting the value of \[P(x)\] from eq. (3) in eq. (5) we will get,
\[ I = {e^{\int {\cot xdx} }} \\
   = {e^{\ln \sin x}} \\
   = \sin x \] …………………………………….. (6)
Now multiplying the integrating factor, \[\sin x\] on both the sides of eq. (6) we will get,
\[ \Rightarrow \sin x\dfrac{{dy}}{{dx}} + \sin x \cdot y\cot x = \sin x \cdot 4x\cos ecx \\
  \Rightarrow \sin x\dfrac{{dy}}{{dx}} + y\cos x = 4x \\
\Rightarrow \sin x\dfrac{{dy}}{{dx}} + y\dfrac{d}{{dx}}\left( {\sin x} \right) = 4x \\ \] ………………………………….. (7)
We know the product rule of differentiation which is given by
\[\dfrac{d}{{dx}}(uv) = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}\] …………………………………….. (8)
Applying this rule to eq. (7), we have,
\[ \Rightarrow \dfrac{d}{{dx}}\left( {y\sin x} \right) = 4x \\
   \Rightarrow d\left( {y\sin x} \right) = 4xdx \] …………………………………………. (9)
Integrating both sides of eq. (9) we will get
\[ \Rightarrow \int {d\left( {y\sin x} \right)} = \int {4xdx} \\
   \Rightarrow y\sin x = 4 \times \dfrac{{{x^2}}}{2} + c \\
  \Rightarrow y\sin x = 2{x^2} + c \] ………………………………………. (10)
As we have to obtain the particular solution of the differential equation we must have the value of integration constant which can be obtained by substituting the respective values of \[x = \dfrac{\pi }{2}\] and \[y = 0\] given in the question in eq. (10), we obtain,
\[ \Rightarrow 0 \cdot \sin \dfrac{\pi }{2} = 2{\left( {\dfrac{\pi }{2}} \right)^2} + c \\
   \Rightarrow c = - \dfrac{{{\pi ^2}}}{2} \\ \] ………………………………………… (11)
Now substituting the value of eq. (11) in eq. (10), we will get the particular solution
\[y\sin x = 2{x^2} - \dfrac{{{\pi ^2}}}{2} \] …………………………………………… (12)
This is the required solution.

Note: While determining the integrating factor, it must be observed that \[P(x)\] must be the function of x only. An alternative method we can solve the differential equation; we can replace \[\cot x\] with \[\dfrac{{\cos x}}{{\sin x}}\] to get the step to eq. (7) and proceeding accordingly.