Solve the differential equation \(\sin x\frac{{dy}}{{dx}} + \frac{y}{

1.	Solve the differential equation \(\sin x\frac{{dy}}{{dx}} + \frac{y}{{\sin x}} = x\sin x{e^{\cot x}}\)1. \(y{e^{ - \cot x}} = \frac{{{x^2}}}{2} + c\)2. \(y{e^{\cot x}} = \frac{{{x^2}}}{2} + c\)3. \(y{e^{ - \cot x}} = x + c\)4. \(y{e^{\cot x}} = x + c\)
Answer» Correct Answer - Option 1 : \(y{e^{ - \cot x}} = \frac{{{x^2}}}{2} + c\) Concept: The standard form of a linear equation of the first order is given by \(\frac{{dy}}{{dx}} + Py = Q\) where P, Q are arbitrary function of x. The integrating factor of the linear equation is given by \(I.F. = {e^{\smallint pdx}}\) The solution of the linear equation is given by \(y\left( {I.F.} \right) = \smallint Q\left( {I.F.} \right)dx + c.\) Calculation: \(\sin x\frac{{dy}}{{dx}} + \frac{y}{{\sin x}} = x\sin x{e^{\cot x}}\) \(\frac{{dy}}{{dx}} + \frac{y}{{{{\sin }^2}x}} = x.{e^{\cot x}}\) It is form of \(\frac{{dy}}{{dx}} + Py = Q\) \(I.F. = {e^{\smallint pdx}}\) \(I.F. = {e^{\smallint cose{c^2}xdx}} = {e^{ - \cot x}}\) The solution of the linear equation is given by \(y\left( {I.F.} \right) = \smallint Q\left( {I.F.} \right)dx + c\) \(y{e^{ - \cot x}} = \smallint x{e^{\cot x}}.{e^{ - \cot x}}dx + c\) \(y{e^{ - \cot x}} = \smallint xdx + c\) \(y{e^{ - \cot x}} = \frac{{{x^2}}}{2} + c\)

Solve the differential equation \(\sin x\frac{{dy}}{{dx}} + \frac{y}{{\sin x}} = x\sin x{e^{\cot x}}\)1. \(y{e^{ - \cot x}} = \frac{{{x^2}}}{2} + c\)2. \(y{e^{\cot x}} = \frac{{{x^2}}}{2} + c\)3. \(y{e^{ - \cot x}} = x + c\)4. \(y{e^{\cot x}} = x + c\)

Answer» Correct Answer - Option 1 : \(y{e^{ - \cot x}} = \frac{{{x^2}}}{2} + c\)

Concept:

The standard form of a linear equation of the first order is given by \(\frac{{dy}}{{dx}} + Py = Q\) where P, Q are arbitrary function of x.

The integrating factor of the linear equation is given by \(I.F. = {e^{\smallint pdx}}\)

The solution of the linear equation is given by \(y\left( {I.F.} \right) = \smallint Q\left( {I.F.} \right)dx + c.\)

Calculation:

\(\sin x\frac{{dy}}{{dx}} + \frac{y}{{\sin x}} = x\sin x{e^{\cot x}}\)

\(\frac{{dy}}{{dx}} + \frac{y}{{{{\sin }^2}x}} = x.{e^{\cot x}}\)

It is form of \(\frac{{dy}}{{dx}} + Py = Q\)

\(I.F. = {e^{\smallint pdx}}\)

\(I.F. = {e^{\smallint cose{c^2}xdx}} = {e^{ - \cot x}}\)

The solution of the linear equation is given by

\(y\left( {I.F.} \right) = \smallint Q\left( {I.F.} \right)dx + c\)

\(y{e^{ - \cot x}} = \smallint x{e^{\cot x}}.{e^{ - \cot x}}dx + c\)

\(y{e^{ - \cot x}} = \smallint xdx + c\)

\(y{e^{ - \cot x}} = \frac{{{x^2}}}{2} + c\)