If \(\frac{{dx}}{{dy}} + \frac{x}{{co{s^2}y}} = {e^{ - \tan y}}\), and

1.	If \(\frac{{dx}}{{dy}} + \frac{x}{{co{s^2}y}} = {e^{ - \tan y}}\), and x(0) = 0, then x (π / 4) is equal to1. \(\frac{\pi }{{4e}}\)2. \(\frac{\pi }{{4e}} - \frac{1}{e}\)3. \(\frac{\pi }{{4e}} + \frac{1}{e}\)4. \(\frac{{4e}}{\pi }\)
Answer» Correct Answer - Option 1 : \(\frac{\pi }{{4e}}\) 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: \(\frac{{dx}}{{dy}} + \frac{x}{{co{s^2}y}} = {e^{ - \tan y}}\) It is the linear equation of the form of \(\frac{{dx}}{{dy}} + Px = Q\) \(I.F. = {e^{\smallint pdy}}\) \(I.F. = {e^{\smallint se{c^2}ydy}} = {e^{\tan y}}\) The solution of the linear equation is given by \(x\left( {I.F.} \right) = \smallint Q\left( {I.F.} \right)dy + c\) \(x{e^{\tan y}} = \smallint {e^{\tan y}}.{e^{ - \tan y}}dy + c\) \(x{e^{tany}} = y + c\) Put x = 0 and y = 0 in the above equation we get, c = 0 \(x = y{e^{ - \tan y}}\) Put y = π / 4 in the above equation we get \(x = \frac{\pi }{4}{e^{ - tan\frac{\pi }{4}}}\) \(x = \frac{\pi }{{4e}}\)

If \(\frac{{dx}}{{dy}} + \frac{x}{{co{s^2}y}} = {e^{ - \tan y}}\), and x(0) = 0, then x (π / 4) is equal to1. \(\frac{\pi }{{4e}}\)2. \(\frac{\pi }{{4e}} - \frac{1}{e}\)3. \(\frac{\pi }{{4e}} + \frac{1}{e}\)4. \(\frac{{4e}}{\pi }\)

Answer» Correct Answer - Option 1 : \(\frac{\pi }{{4e}}\)

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:

\(\frac{{dx}}{{dy}} + \frac{x}{{co{s^2}y}} = {e^{ - \tan y}}\)

It is the linear equation of the form of \(\frac{{dx}}{{dy}} + Px = Q\)

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

\(I.F. = {e^{\smallint se{c^2}ydy}} = {e^{\tan y}}\)

The solution of the linear equation is given by

\(x\left( {I.F.} \right) = \smallint Q\left( {I.F.} \right)dy + c\)

\(x{e^{\tan y}} = \smallint {e^{\tan y}}.{e^{ - \tan y}}dy + c\)

\(x{e^{tany}} = y + c\)

Put x = 0 and y = 0 in the above equation we get, c = 0

\(x = y{e^{ - \tan y}}\)

Put y = π / 4 in the above equation we get

\(x = \frac{\pi }{4}{e^{ - tan\frac{\pi }{4}}}\)

\(x = \frac{\pi }{{4e}}\)