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Differential Equation

87562 Solution of the differential equation
$x = 1 + x y \frac{d y}{d x} + \frac{(x y)^{2}}{2!} {(\frac{d y}{d x})}^{2} + \frac{(x y)^{3}}{3!} {(\frac{d y}{d x})}^{3} + \dots . . i s$

1

y = \log_{e} (x) + c

2

y = {(\log_{e} x)}^{2} + c

3

y = \pm \sqrt{{(\log_{e} x)}^{2} + 2 c}

4

x y = x^{y} + k

Explanation:

(C) : The given differential equation can be written as-
$x = e^{x y \frac{d y}{d x}}$
$\log x = x y \frac{d y}{d x}$
$y d y = \frac{\log x}{x} d x$
$y d y = \log x d (\log x)$
On integrating, we get-
$\frac{y^{2}}{2} = \frac{(\log x)^{2}}{2} + c$
$y^{2} = {(\log_{e} x)}^{2} + 2 c$
$y = \pm \sqrt{{(\log_{e} x)}^{2} + 2 c}$

Manipal UGET-2010

Differential Equation

87563 The solution of the differential equation $\frac{d y}{d x} = \sin (x + y) \tan (x + y) - 1$ is

1

cosec (x + y) + \tan (x + y) = x + C

2

x + cosec (x + y) = C

3

x + \tan (x + y) = C

4

x + \sec (x + y) = C

Explanation:

(B) : We have,
$\frac{d y}{d x} = \sin (x + y) \tan (x + y) - 1$
Put, $x + y = v$ , we get-
$1 + \frac{d y}{d x} = \frac{d v}{d x}$
$\frac{d y}{d x} = \frac{d v}{d x} - 1$
Equation (i) reducing to,
$\frac{dv}{dx} = \sin v \tan v$
$\frac{d v}{\sin v \tan v} = d x$
$\int cosec v \cot v d v = \int dx$
$- cosec v cotv = x + C$
$- cosec (x + y) = x + C$
$x + cosec (x + y) = C$

Manipal UGET-2016

Differential Equation

87564 The solution of differential equation
$(y \log x - 1) y d x = x d y$ is

1

y (\log e^{x} + C x) = 1

2

(\log \frac{x}{e} + C x) x = y

3

(\log C x^{2} + e x^{2}) y = x

4 None of these

Explanation:

(D) : Given differential equation is:-
$(y \log x - 1) y d x = x d y$
$\frac{dy}{dx} = \frac{(y \log x - 1) y}{x}$
$\frac{d y}{d x} = \frac{y^{2} \log x}{x} - \frac{y}{x}$
$\frac{d y}{d x} + \frac{y}{x} = \frac{y^{2} \log x}{x}$
$\frac{1}{y^{2}} \frac{d y}{d x} + \frac{y^{- 1}}{x} = \frac{\log x}{x}$
Put,
$y^{- 1} = v$
$- y^{- 2} \frac{d y}{d x} = \frac{d v}{d x}$
$y^{- 2} \frac{d y}{d x} = - \frac{d v}{d x}$
From ${Eq}^{n}$ (i), we have:-
$- \frac{d v}{d x} + \frac{v}{x} = \frac{\log x}{x} \Rightarrow \frac{d y}{d x} - \frac{v}{x} = - \frac{\log x}{x}$
Here
$P = \frac{- 1}{x}, Q = - \frac{\log x}{x}$
$I \cdot F \cdot = e^{\int - \frac{1}{x} d x} = e^{- \log x} = \frac{1}{x}$
So,
$v \cdot \frac{1}{x} = \int \frac{1}{x} (- \frac{\log x}{x}) dx + C$
$= - [\log x (\frac{- 1}{x}) + \int \frac{1}{x} \cdot \frac{1}{x} dx] + C$
$\frac{1}{x \cdot y} = \frac{\log x}{x} + \frac{1}{x} + C [∵ v = \frac{1}{y}]$
$1 = y [\log x + 1 + Cx]$
$1 = y [\log x + \log e + Cx]$
$1 = y [\log (e \cdot x) + cx]$

Manipal UGET-2020

Differential Equation

87565 Solution of the equation $\cos^{2} x \frac{d y}{d x} - (\tan 2 x) y = \cos^{4} x, | x | < \frac{π}{4}$ , where $y (\frac{π}{6}) = \frac{3 \sqrt{3}}{8}$ , is given by

1

y \frac{\tan 2 x}{1 - \tan^{2} x} = 0

2

y (1 - \tan^{2} x) = C

3

y = \sin 2 x + C

4

y = \frac{1}{2} \cdot \frac{\sin 2 x}{1 - \tan^{2} x}

Explanation:

(D) :
$\cos^{2} x \frac{d y}{d x} - (\tan 2 x) y = \cos^{4} x$
Similar as:-
$\frac{d y}{d x} - (\frac{\tan 2 x}{\cos^{2} x}) y = \cos^{2} x$
$\frac{d y}{d x} + p y = θ$
$I.F. = e^{\int P d x}$
Now,
$\int P d x = - \int \frac{\tan 2 x}{\cos^{2} x} d x$
$∴ = - \int \frac{2 \sin 2 x}{\cos 2 x (1 + \cos 2 x)}$
Let, $\cos 2 x = t$
$- 2 \sin 2 x d x = d t$
So, $\int P d x = \int \frac{d t}{t (t + 1)}$
$= \int \frac{t + 1 - t}{t (t + 1)} dt = \int \frac{dt}{t} - \int \frac{dt}{t + 1}$
$= \ln | t | - \ln | t + 1 |$
$= \ln | \frac{t}{t + 1} | = \ln | \frac{\cos 2 x}{1 + \cos 2 x} |$
So, $\ln | \frac{\cos 2 x}{1 + \cos 2 x} |$
I.F $= e \frac{\cos 2 x}{1 + \cos 2 x}$
Now, solution is:-
$Y \cdot I \cdot F \cdot = \int I \cdot F \cdot x θ dx + c$
$y \times \frac{\cos 2 x}{1 + \cos 2 x} = \int \frac{\cos 2 x}{1 + \cos 2 x} \times \cos^{2} xdx + C$
$= \int \frac{\cos 2 x}{2 \cos^{2} x} \times \cos^{2} x d x + C$
$= \frac{1}{2} \int \cos 2 xdx + C = \frac{1}{2} \times \frac{\sin 2 x}{2} + C$
$\Rightarrow y (\frac{\cos 2 x}{1 + \cos 2 x}) = \frac{1}{2} \frac{\sin 2 x}{2} + C$
Now,
$y (\frac{π}{6}) = \frac{3 \sqrt{3}}{8}$ means:-
at $x = \frac{π}{6}, y = \frac{3 \sqrt{3}}{8}$
So, from (i) :-
$\frac{3 \sqrt{3}}{8} \times \frac{1 / 2}{1 + 1 / 2} = \frac{\sqrt{3}}{2 \times 4} + C$
$\frac{3 \sqrt{3}}{8} \times (\frac{1}{3}) = \frac{\sqrt{3}}{8} + C$
$C = \frac{\sqrt{3}}{8} - \frac{\sqrt{3}}{8}$
$C = 0$
Now,
${Eq}^{n}$ (i) be comes:-
$y (\frac{\cos 2 x}{\cos 2 x + 1}) = \frac{1}{4} \sin 2 x + 0$
$y = \frac{\frac{1}{4} \sin 2 x}{\frac{\cos 2 x}{\cos 2 x + 1}} = \frac{1}{4} \frac{\sin 2 x}{\frac{\cos^{2} x - \sin^{2} x}{2 \cos^{2} x}}$
$y = \frac{1}{2} \cdot \frac{\sin 2 x}{1 - \tan^{2} x}$

Differential Equation

87562 Solution of the differential equation
$x = 1 + x y \frac{d y}{d x} + \frac{(x y)^{2}}{2!} {(\frac{d y}{d x})}^{2} + \frac{(x y)^{3}}{3!} {(\frac{d y}{d x})}^{3} + \dots . . i s$

1

y = \log_{e} (x) + c

2

y = {(\log_{e} x)}^{2} + c

3

y = \pm \sqrt{{(\log_{e} x)}^{2} + 2 c}

4

x y = x^{y} + k

Explanation:

(C) : The given differential equation can be written as-
$x = e^{x y \frac{d y}{d x}}$
$\log x = x y \frac{d y}{d x}$
$y d y = \frac{\log x}{x} d x$
$y d y = \log x d (\log x)$
On integrating, we get-
$\frac{y^{2}}{2} = \frac{(\log x)^{2}}{2} + c$
$y^{2} = {(\log_{e} x)}^{2} + 2 c$
$y = \pm \sqrt{{(\log_{e} x)}^{2} + 2 c}$

Manipal UGET-2010

Differential Equation

87563 The solution of the differential equation $\frac{d y}{d x} = \sin (x + y) \tan (x + y) - 1$ is

1

cosec (x + y) + \tan (x + y) = x + C

2

x + cosec (x + y) = C

3

x + \tan (x + y) = C

4

x + \sec (x + y) = C

Explanation:

(B) : We have,
$\frac{d y}{d x} = \sin (x + y) \tan (x + y) - 1$
Put, $x + y = v$ , we get-
$1 + \frac{d y}{d x} = \frac{d v}{d x}$
$\frac{d y}{d x} = \frac{d v}{d x} - 1$
Equation (i) reducing to,
$\frac{dv}{dx} = \sin v \tan v$
$\frac{d v}{\sin v \tan v} = d x$
$\int cosec v \cot v d v = \int dx$
$- cosec v cotv = x + C$
$- cosec (x + y) = x + C$
$x + cosec (x + y) = C$

Manipal UGET-2016

Differential Equation

87564 The solution of differential equation
$(y \log x - 1) y d x = x d y$ is

1

y (\log e^{x} + C x) = 1

2

(\log \frac{x}{e} + C x) x = y

3

(\log C x^{2} + e x^{2}) y = x

4 None of these

Explanation:

(D) : Given differential equation is:-
$(y \log x - 1) y d x = x d y$
$\frac{dy}{dx} = \frac{(y \log x - 1) y}{x}$
$\frac{d y}{d x} = \frac{y^{2} \log x}{x} - \frac{y}{x}$
$\frac{d y}{d x} + \frac{y}{x} = \frac{y^{2} \log x}{x}$
$\frac{1}{y^{2}} \frac{d y}{d x} + \frac{y^{- 1}}{x} = \frac{\log x}{x}$
Put,
$y^{- 1} = v$
$- y^{- 2} \frac{d y}{d x} = \frac{d v}{d x}$
$y^{- 2} \frac{d y}{d x} = - \frac{d v}{d x}$
From ${Eq}^{n}$ (i), we have:-
$- \frac{d v}{d x} + \frac{v}{x} = \frac{\log x}{x} \Rightarrow \frac{d y}{d x} - \frac{v}{x} = - \frac{\log x}{x}$
Here
$P = \frac{- 1}{x}, Q = - \frac{\log x}{x}$
$I \cdot F \cdot = e^{\int - \frac{1}{x} d x} = e^{- \log x} = \frac{1}{x}$
So,
$v \cdot \frac{1}{x} = \int \frac{1}{x} (- \frac{\log x}{x}) dx + C$
$= - [\log x (\frac{- 1}{x}) + \int \frac{1}{x} \cdot \frac{1}{x} dx] + C$
$\frac{1}{x \cdot y} = \frac{\log x}{x} + \frac{1}{x} + C [∵ v = \frac{1}{y}]$
$1 = y [\log x + 1 + Cx]$
$1 = y [\log x + \log e + Cx]$
$1 = y [\log (e \cdot x) + cx]$

Manipal UGET-2020

Differential Equation

87565 Solution of the equation $\cos^{2} x \frac{d y}{d x} - (\tan 2 x) y = \cos^{4} x, | x | < \frac{π}{4}$ , where $y (\frac{π}{6}) = \frac{3 \sqrt{3}}{8}$ , is given by

1

y \frac{\tan 2 x}{1 - \tan^{2} x} = 0

2

y (1 - \tan^{2} x) = C

3

y = \sin 2 x + C

4

y = \frac{1}{2} \cdot \frac{\sin 2 x}{1 - \tan^{2} x}

Explanation:

(D) :
$\cos^{2} x \frac{d y}{d x} - (\tan 2 x) y = \cos^{4} x$
Similar as:-
$\frac{d y}{d x} - (\frac{\tan 2 x}{\cos^{2} x}) y = \cos^{2} x$
$\frac{d y}{d x} + p y = θ$
$I.F. = e^{\int P d x}$
Now,
$\int P d x = - \int \frac{\tan 2 x}{\cos^{2} x} d x$
$∴ = - \int \frac{2 \sin 2 x}{\cos 2 x (1 + \cos 2 x)}$
Let, $\cos 2 x = t$
$- 2 \sin 2 x d x = d t$
So, $\int P d x = \int \frac{d t}{t (t + 1)}$
$= \int \frac{t + 1 - t}{t (t + 1)} dt = \int \frac{dt}{t} - \int \frac{dt}{t + 1}$
$= \ln | t | - \ln | t + 1 |$
$= \ln | \frac{t}{t + 1} | = \ln | \frac{\cos 2 x}{1 + \cos 2 x} |$
So, $\ln | \frac{\cos 2 x}{1 + \cos 2 x} |$
I.F $= e \frac{\cos 2 x}{1 + \cos 2 x}$
Now, solution is:-
$Y \cdot I \cdot F \cdot = \int I \cdot F \cdot x θ dx + c$
$y \times \frac{\cos 2 x}{1 + \cos 2 x} = \int \frac{\cos 2 x}{1 + \cos 2 x} \times \cos^{2} xdx + C$
$= \int \frac{\cos 2 x}{2 \cos^{2} x} \times \cos^{2} x d x + C$
$= \frac{1}{2} \int \cos 2 xdx + C = \frac{1}{2} \times \frac{\sin 2 x}{2} + C$
$\Rightarrow y (\frac{\cos 2 x}{1 + \cos 2 x}) = \frac{1}{2} \frac{\sin 2 x}{2} + C$
Now,
$y (\frac{π}{6}) = \frac{3 \sqrt{3}}{8}$ means:-
at $x = \frac{π}{6}, y = \frac{3 \sqrt{3}}{8}$
So, from (i) :-
$\frac{3 \sqrt{3}}{8} \times \frac{1 / 2}{1 + 1 / 2} = \frac{\sqrt{3}}{2 \times 4} + C$
$\frac{3 \sqrt{3}}{8} \times (\frac{1}{3}) = \frac{\sqrt{3}}{8} + C$
$C = \frac{\sqrt{3}}{8} - \frac{\sqrt{3}}{8}$
$C = 0$
Now,
${Eq}^{n}$ (i) be comes:-
$y (\frac{\cos 2 x}{\cos 2 x + 1}) = \frac{1}{4} \sin 2 x + 0$
$y = \frac{\frac{1}{4} \sin 2 x}{\frac{\cos 2 x}{\cos 2 x + 1}} = \frac{1}{4} \frac{\sin 2 x}{\frac{\cos^{2} x - \sin^{2} x}{2 \cos^{2} x}}$
$y = \frac{1}{2} \cdot \frac{\sin 2 x}{1 - \tan^{2} x}$

Differential Equation

87562 Solution of the differential equation
$x = 1 + x y \frac{d y}{d x} + \frac{(x y)^{2}}{2!} {(\frac{d y}{d x})}^{2} + \frac{(x y)^{3}}{3!} {(\frac{d y}{d x})}^{3} + \dots . . i s$

1

y = \log_{e} (x) + c

2

y = {(\log_{e} x)}^{2} + c

3

y = \pm \sqrt{{(\log_{e} x)}^{2} + 2 c}

4

x y = x^{y} + k

Explanation:

(C) : The given differential equation can be written as-
$x = e^{x y \frac{d y}{d x}}$
$\log x = x y \frac{d y}{d x}$
$y d y = \frac{\log x}{x} d x$
$y d y = \log x d (\log x)$
On integrating, we get-
$\frac{y^{2}}{2} = \frac{(\log x)^{2}}{2} + c$
$y^{2} = {(\log_{e} x)}^{2} + 2 c$
$y = \pm \sqrt{{(\log_{e} x)}^{2} + 2 c}$

Manipal UGET-2010

Differential Equation

87563 The solution of the differential equation $\frac{d y}{d x} = \sin (x + y) \tan (x + y) - 1$ is

1

cosec (x + y) + \tan (x + y) = x + C

2

x + cosec (x + y) = C

3

x + \tan (x + y) = C

4

x + \sec (x + y) = C

Explanation:

(B) : We have,
$\frac{d y}{d x} = \sin (x + y) \tan (x + y) - 1$
Put, $x + y = v$ , we get-
$1 + \frac{d y}{d x} = \frac{d v}{d x}$
$\frac{d y}{d x} = \frac{d v}{d x} - 1$
Equation (i) reducing to,
$\frac{dv}{dx} = \sin v \tan v$
$\frac{d v}{\sin v \tan v} = d x$
$\int cosec v \cot v d v = \int dx$
$- cosec v cotv = x + C$
$- cosec (x + y) = x + C$
$x + cosec (x + y) = C$

Manipal UGET-2016

Differential Equation

87564 The solution of differential equation
$(y \log x - 1) y d x = x d y$ is

1

y (\log e^{x} + C x) = 1

2

(\log \frac{x}{e} + C x) x = y

3

(\log C x^{2} + e x^{2}) y = x

4 None of these

Explanation:

(D) : Given differential equation is:-
$(y \log x - 1) y d x = x d y$
$\frac{dy}{dx} = \frac{(y \log x - 1) y}{x}$
$\frac{d y}{d x} = \frac{y^{2} \log x}{x} - \frac{y}{x}$
$\frac{d y}{d x} + \frac{y}{x} = \frac{y^{2} \log x}{x}$
$\frac{1}{y^{2}} \frac{d y}{d x} + \frac{y^{- 1}}{x} = \frac{\log x}{x}$
Put,
$y^{- 1} = v$
$- y^{- 2} \frac{d y}{d x} = \frac{d v}{d x}$
$y^{- 2} \frac{d y}{d x} = - \frac{d v}{d x}$
From ${Eq}^{n}$ (i), we have:-
$- \frac{d v}{d x} + \frac{v}{x} = \frac{\log x}{x} \Rightarrow \frac{d y}{d x} - \frac{v}{x} = - \frac{\log x}{x}$
Here
$P = \frac{- 1}{x}, Q = - \frac{\log x}{x}$
$I \cdot F \cdot = e^{\int - \frac{1}{x} d x} = e^{- \log x} = \frac{1}{x}$
So,
$v \cdot \frac{1}{x} = \int \frac{1}{x} (- \frac{\log x}{x}) dx + C$
$= - [\log x (\frac{- 1}{x}) + \int \frac{1}{x} \cdot \frac{1}{x} dx] + C$
$\frac{1}{x \cdot y} = \frac{\log x}{x} + \frac{1}{x} + C [∵ v = \frac{1}{y}]$
$1 = y [\log x + 1 + Cx]$
$1 = y [\log x + \log e + Cx]$
$1 = y [\log (e \cdot x) + cx]$

Manipal UGET-2020

Differential Equation

87565 Solution of the equation $\cos^{2} x \frac{d y}{d x} - (\tan 2 x) y = \cos^{4} x, | x | < \frac{π}{4}$ , where $y (\frac{π}{6}) = \frac{3 \sqrt{3}}{8}$ , is given by

1

y \frac{\tan 2 x}{1 - \tan^{2} x} = 0

2

y (1 - \tan^{2} x) = C

3

y = \sin 2 x + C

4

y = \frac{1}{2} \cdot \frac{\sin 2 x}{1 - \tan^{2} x}

Explanation:

(D) :
$\cos^{2} x \frac{d y}{d x} - (\tan 2 x) y = \cos^{4} x$
Similar as:-
$\frac{d y}{d x} - (\frac{\tan 2 x}{\cos^{2} x}) y = \cos^{2} x$
$\frac{d y}{d x} + p y = θ$
$I.F. = e^{\int P d x}$
Now,
$\int P d x = - \int \frac{\tan 2 x}{\cos^{2} x} d x$
$∴ = - \int \frac{2 \sin 2 x}{\cos 2 x (1 + \cos 2 x)}$
Let, $\cos 2 x = t$
$- 2 \sin 2 x d x = d t$
So, $\int P d x = \int \frac{d t}{t (t + 1)}$
$= \int \frac{t + 1 - t}{t (t + 1)} dt = \int \frac{dt}{t} - \int \frac{dt}{t + 1}$
$= \ln | t | - \ln | t + 1 |$
$= \ln | \frac{t}{t + 1} | = \ln | \frac{\cos 2 x}{1 + \cos 2 x} |$
So, $\ln | \frac{\cos 2 x}{1 + \cos 2 x} |$
I.F $= e \frac{\cos 2 x}{1 + \cos 2 x}$
Now, solution is:-
$Y \cdot I \cdot F \cdot = \int I \cdot F \cdot x θ dx + c$
$y \times \frac{\cos 2 x}{1 + \cos 2 x} = \int \frac{\cos 2 x}{1 + \cos 2 x} \times \cos^{2} xdx + C$
$= \int \frac{\cos 2 x}{2 \cos^{2} x} \times \cos^{2} x d x + C$
$= \frac{1}{2} \int \cos 2 xdx + C = \frac{1}{2} \times \frac{\sin 2 x}{2} + C$
$\Rightarrow y (\frac{\cos 2 x}{1 + \cos 2 x}) = \frac{1}{2} \frac{\sin 2 x}{2} + C$
Now,
$y (\frac{π}{6}) = \frac{3 \sqrt{3}}{8}$ means:-
at $x = \frac{π}{6}, y = \frac{3 \sqrt{3}}{8}$
So, from (i) :-
$\frac{3 \sqrt{3}}{8} \times \frac{1 / 2}{1 + 1 / 2} = \frac{\sqrt{3}}{2 \times 4} + C$
$\frac{3 \sqrt{3}}{8} \times (\frac{1}{3}) = \frac{\sqrt{3}}{8} + C$
$C = \frac{\sqrt{3}}{8} - \frac{\sqrt{3}}{8}$
$C = 0$
Now,
${Eq}^{n}$ (i) be comes:-
$y (\frac{\cos 2 x}{\cos 2 x + 1}) = \frac{1}{4} \sin 2 x + 0$
$y = \frac{\frac{1}{4} \sin 2 x}{\frac{\cos 2 x}{\cos 2 x + 1}} = \frac{1}{4} \frac{\sin 2 x}{\frac{\cos^{2} x - \sin^{2} x}{2 \cos^{2} x}}$
$y = \frac{1}{2} \cdot \frac{\sin 2 x}{1 - \tan^{2} x}$

Differential Equation

87562 Solution of the differential equation
$x = 1 + x y \frac{d y}{d x} + \frac{(x y)^{2}}{2!} {(\frac{d y}{d x})}^{2} + \frac{(x y)^{3}}{3!} {(\frac{d y}{d x})}^{3} + \dots . . i s$

1

y = \log_{e} (x) + c

2

y = {(\log_{e} x)}^{2} + c

3

y = \pm \sqrt{{(\log_{e} x)}^{2} + 2 c}

4

x y = x^{y} + k

Explanation:

(C) : The given differential equation can be written as-
$x = e^{x y \frac{d y}{d x}}$
$\log x = x y \frac{d y}{d x}$
$y d y = \frac{\log x}{x} d x$
$y d y = \log x d (\log x)$
On integrating, we get-
$\frac{y^{2}}{2} = \frac{(\log x)^{2}}{2} + c$
$y^{2} = {(\log_{e} x)}^{2} + 2 c$
$y = \pm \sqrt{{(\log_{e} x)}^{2} + 2 c}$

Manipal UGET-2010

Differential Equation

87563 The solution of the differential equation $\frac{d y}{d x} = \sin (x + y) \tan (x + y) - 1$ is

1

cosec (x + y) + \tan (x + y) = x + C

2

x + cosec (x + y) = C

3

x + \tan (x + y) = C

4

x + \sec (x + y) = C

Explanation:

(B) : We have,
$\frac{d y}{d x} = \sin (x + y) \tan (x + y) - 1$
Put, $x + y = v$ , we get-
$1 + \frac{d y}{d x} = \frac{d v}{d x}$
$\frac{d y}{d x} = \frac{d v}{d x} - 1$
Equation (i) reducing to,
$\frac{dv}{dx} = \sin v \tan v$
$\frac{d v}{\sin v \tan v} = d x$
$\int cosec v \cot v d v = \int dx$
$- cosec v cotv = x + C$
$- cosec (x + y) = x + C$
$x + cosec (x + y) = C$

Manipal UGET-2016

Differential Equation

87564 The solution of differential equation
$(y \log x - 1) y d x = x d y$ is

1

y (\log e^{x} + C x) = 1

2

(\log \frac{x}{e} + C x) x = y

3

(\log C x^{2} + e x^{2}) y = x

4 None of these

Explanation:

(D) : Given differential equation is:-
$(y \log x - 1) y d x = x d y$
$\frac{dy}{dx} = \frac{(y \log x - 1) y}{x}$
$\frac{d y}{d x} = \frac{y^{2} \log x}{x} - \frac{y}{x}$
$\frac{d y}{d x} + \frac{y}{x} = \frac{y^{2} \log x}{x}$
$\frac{1}{y^{2}} \frac{d y}{d x} + \frac{y^{- 1}}{x} = \frac{\log x}{x}$
Put,
$y^{- 1} = v$
$- y^{- 2} \frac{d y}{d x} = \frac{d v}{d x}$
$y^{- 2} \frac{d y}{d x} = - \frac{d v}{d x}$
From ${Eq}^{n}$ (i), we have:-
$- \frac{d v}{d x} + \frac{v}{x} = \frac{\log x}{x} \Rightarrow \frac{d y}{d x} - \frac{v}{x} = - \frac{\log x}{x}$
Here
$P = \frac{- 1}{x}, Q = - \frac{\log x}{x}$
$I \cdot F \cdot = e^{\int - \frac{1}{x} d x} = e^{- \log x} = \frac{1}{x}$
So,
$v \cdot \frac{1}{x} = \int \frac{1}{x} (- \frac{\log x}{x}) dx + C$
$= - [\log x (\frac{- 1}{x}) + \int \frac{1}{x} \cdot \frac{1}{x} dx] + C$
$\frac{1}{x \cdot y} = \frac{\log x}{x} + \frac{1}{x} + C [∵ v = \frac{1}{y}]$
$1 = y [\log x + 1 + Cx]$
$1 = y [\log x + \log e + Cx]$
$1 = y [\log (e \cdot x) + cx]$

Manipal UGET-2020

Differential Equation

87565 Solution of the equation $\cos^{2} x \frac{d y}{d x} - (\tan 2 x) y = \cos^{4} x, | x | < \frac{π}{4}$ , where $y (\frac{π}{6}) = \frac{3 \sqrt{3}}{8}$ , is given by

1

y \frac{\tan 2 x}{1 - \tan^{2} x} = 0

2

y (1 - \tan^{2} x) = C

3

y = \sin 2 x + C

4

y = \frac{1}{2} \cdot \frac{\sin 2 x}{1 - \tan^{2} x}

Explanation:

(D) :
$\cos^{2} x \frac{d y}{d x} - (\tan 2 x) y = \cos^{4} x$
Similar as:-
$\frac{d y}{d x} - (\frac{\tan 2 x}{\cos^{2} x}) y = \cos^{2} x$
$\frac{d y}{d x} + p y = θ$
$I.F. = e^{\int P d x}$
Now,
$\int P d x = - \int \frac{\tan 2 x}{\cos^{2} x} d x$
$∴ = - \int \frac{2 \sin 2 x}{\cos 2 x (1 + \cos 2 x)}$
Let, $\cos 2 x = t$
$- 2 \sin 2 x d x = d t$
So, $\int P d x = \int \frac{d t}{t (t + 1)}$
$= \int \frac{t + 1 - t}{t (t + 1)} dt = \int \frac{dt}{t} - \int \frac{dt}{t + 1}$
$= \ln | t | - \ln | t + 1 |$
$= \ln | \frac{t}{t + 1} | = \ln | \frac{\cos 2 x}{1 + \cos 2 x} |$
So, $\ln | \frac{\cos 2 x}{1 + \cos 2 x} |$
I.F $= e \frac{\cos 2 x}{1 + \cos 2 x}$
Now, solution is:-
$Y \cdot I \cdot F \cdot = \int I \cdot F \cdot x θ dx + c$
$y \times \frac{\cos 2 x}{1 + \cos 2 x} = \int \frac{\cos 2 x}{1 + \cos 2 x} \times \cos^{2} xdx + C$
$= \int \frac{\cos 2 x}{2 \cos^{2} x} \times \cos^{2} x d x + C$
$= \frac{1}{2} \int \cos 2 xdx + C = \frac{1}{2} \times \frac{\sin 2 x}{2} + C$
$\Rightarrow y (\frac{\cos 2 x}{1 + \cos 2 x}) = \frac{1}{2} \frac{\sin 2 x}{2} + C$
Now,
$y (\frac{π}{6}) = \frac{3 \sqrt{3}}{8}$ means:-
at $x = \frac{π}{6}, y = \frac{3 \sqrt{3}}{8}$
So, from (i) :-
$\frac{3 \sqrt{3}}{8} \times \frac{1 / 2}{1 + 1 / 2} = \frac{\sqrt{3}}{2 \times 4} + C$
$\frac{3 \sqrt{3}}{8} \times (\frac{1}{3}) = \frac{\sqrt{3}}{8} + C$
$C = \frac{\sqrt{3}}{8} - \frac{\sqrt{3}}{8}$
$C = 0$
Now,
${Eq}^{n}$ (i) be comes:-
$y (\frac{\cos 2 x}{\cos 2 x + 1}) = \frac{1}{4} \sin 2 x + 0$
$y = \frac{\frac{1}{4} \sin 2 x}{\frac{\cos 2 x}{\cos 2 x + 1}} = \frac{1}{4} \frac{\sin 2 x}{\frac{\cos^{2} x - \sin^{2} x}{2 \cos^{2} x}}$
$y = \frac{1}{2} \cdot \frac{\sin 2 x}{1 - \tan^{2} x}$

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