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Question: The solution of \(\dfrac{{dy}}{{dx}} - 3y\cot x = \sin 2x\) is: A. \(y + {\sin ^2}x = c{\sin ^3}x\...

The solution of dydx−3ycot⁡x=sin⁡2x\dfrac{{dy}}{{dx}} - 3y\cot x = \sin 2x is:
A. y+sin⁡2x=csin⁡3xy + {\sin ^2}x = c{\sin ^3}x
B. y+sin⁡x=c(sin⁡3x+cos⁡x)y + \sin x = c\left( {{{\sin }^3}x + \cos x} \right)
C. y+sin⁡2x=sin⁡3xy + {\sin ^2}x = {\sin ^3}x
D. y=−2sin⁡2x+csin⁡3xy = - 2{\sin ^2}x + c{\sin ^3}x

Explanation

Solution

Hint: We will begin the solution by comparing the given equation with the general form, dydx+Py=Q\dfrac{{dy}}{{dx}} + Py = Q and find the values of P and Q. Once we have determined the values of P and Q, then we will be using the formula for the Integrating factor, which is given by, IF=e∫Pdx{\text{IF}} = {e^{\int {Pdx} }} and substitute the value of P to calculate it. In the end, when we have calculated or determined P, Q, and the integrating factor, we will be using the formula of the solution of the differential equation, given by, y×IF=∫Q⋅IFdxy \times {\text{IF}} = \int {Q \cdot {\text{IF}}dx} to find the required solution.

Complete step-by-step answer
Let us consider the given equation,
dydx−3ycot⁡x=sin⁡2x\dfrac{{dy}}{{dx}} - 3y\cot x = \sin 2x

This equation can be compared with the standard form, dydx+Py=Q\dfrac{{dy}}{{dx}} + Py = Q to find the value of P and Q.

Thus, on comparing we get;
P=−3cot⁡xP = - 3\cot x
And,
Q=sin⁡2xQ = \sin 2x

Now, we will now be calculating the integrating factor, using the formula;
IF=e∫Pdx{\text{IF}} = {e^{\int {Pdx} }}

We will now substitute, P=−3cot⁡xP = - 3\cot x into the above formula, and find the value of the integrating factor.

⇒e∫−3cot⁡xdx=e−3∫cot⁡xdx =e−3log⁡∣sin⁡x∣ =elog⁡∣sin⁡x∣−3 =elog⁡1∣sin⁡3x∣ =elog⁡∣cos⁡ec3x∣ =cos⁡ec3x  \Rightarrow {e^{\int { - 3\cot xdx} }} = {e^{ - 3\int {\cot xdx} }} \\\ = {e^{ - 3\log |\sin x|}} \\\ = {e^{\log |\sin x{|^{ - 3}}}} \\\ = {e^{\log \dfrac{1}{{|{{\sin }^3}x|}}}} \\\ = {e^{\log |\cos e{c^3}x|}} \\\ = \cos e{c^3}x \\\

Thus, we get the value of the integrating factor to be equal to cos⁡ec3x\cos e{c^3}x.

Now, we know that the solution to the differential equation of the given form, is given by;
y×IF=∫Q⋅IFdxy \times {\text{IF}} = \int {Q \cdot {\text{IF}}dx} . Thus, we substitute the given values into the formula to find the solution to the given equation.

y×cos⁡ec3x=∫sin⁡2x⋅cos⁡ec3xdx ⇒y×cos⁡ec3x=∫2sin⁡xcos⁡xsin⁡3xdx ⇒ycos⁡ec3x=∫2cos⁡xsin⁡2xdx ⇒ycos⁡ec3x=∫2cos⁡xsin⁡x×1sin⁡xdx ⇒ycos⁡ec3x=2∫cot⁡xcos⁡ecxdx  y \times \cos e{c^3}x = \int {\sin 2x \cdot \cos e{c^3}xdx} \\\ \Rightarrow y \times \cos e{c^3}x = \int {\dfrac{{2\sin x\cos x}}{{{{\sin }^3}x}}dx} \\\ \Rightarrow y\cos e{c^3}x = \int {\dfrac{{2\cos x}}{{{{\sin }^2}x}}dx} \\\ \Rightarrow y\cos e{c^3}x = \int {2\dfrac{{\cos x}}{{\sin x}} \times \dfrac{1}{{\sin x}}dx} \\\ \Rightarrow y\cos e{c^3}x = 2\int {\cot x\cos ecxdx} \\\

Now, we will use the known integration that, ∫cot⁡xcos⁡ecxdx=−cos⁡ecx+c\int {\cot x\cos ecxdx = - \cos ecx + c} .

⇒ycos⁡ec3x=2(−cos⁡ecx)+c ⇒y=−2cos⁡ecxcos⁡ec3x+ccos⁡ec3x ⇒y=−2cos⁡ec2x+ccos⁡ec3x ⇒y=−2sin⁡2x+csin⁡3x  \Rightarrow y\cos e{c^3}x = 2\left( { - \cos ecx} \right) + c \\\ \Rightarrow y = \dfrac{{ - 2\cos ecx}}{{\cos e{c^3}x}} + \dfrac{c}{{\cos e{c^3}x}} \\\ \Rightarrow y = \dfrac{{ - 2}}{{\cos e{c^2}x}} + \dfrac{c}{{\cos e{c^3}x}} \\\ \Rightarrow y = - 2{\sin ^2}x + c{\sin ^3}x \\\

which is the required solution.

Hence, option (D) is the correct option.

Note: There are numerous ways to solve these types of questions, like, for example, you could have begun the solution using the trigonometric relations and formulae and tried finding the solution to the given equation by separating the variables. But after a certain number of steps you would have noticed that the complexity of the equation has increased. So, wherever you find equations having the term, dydx\dfrac{{dy}}{{dx}} and the question is asking to find its solutions, just use the method of integrating factor. This method is time saving and reduces the complexity of the given equation.