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i) Integrating factor of the differential of the form \frac{d x}{d y}+P_{1} x=Q_{1} is given by e^{\int P_{1}d y}

(ii) Solution of the differential equation of the type \frac{d x}{d y^{\prime}}+P_{1} x=Q_{1}  is given by x e^{\int P_{1} d y}=\int Q_{1} e^{\int P_{1} d y} d y+C

iii) Correct substitution for the solution of the differential equation of the type \frac{d y}{d x} f(x, y), where f(x, y) is a homogeneous function of zero degree is  y=v x 
(iv) Correct substitution for the solution of the differential equation of the type \frac{d x}{d y} g(x, y)where g(x, y) is a homogeneous function of the degree zero is x=v y
(V) Number of arbitrary constants in the particular solution of a differential
equation of order two is two.

(vi)The differential equation representing the family of circles x^{2}+(y-a)^{2}= a^{2} will be of order two.  
(vii) The solution of \frac{d y}{d x}=\left(\frac{y}{x}\right)^{1 / 3} is y^{\frac{2}{3}}-x^{\frac{2}{3}}=c
 
(viii) Differential equation representing the family of curves y=e^{x}(A \cos x+B \sin x ) is \frac{d^{2} y}{d x^{2}}-2 \frac{d y}{d x}+2 y=0

ix) The solution of the differential equation \frac{d y}{d x}=\frac{x+2 y}{x} is x+y=k x^{2}.

 

x) Solution of  \frac{x d y}{d x}=y+x \tan \left(\frac{y}{x}\right) \text { is } \sin \frac{y}{x}=C x

xi) The differential equation of all non horizontal lines in a plane is  \begin{aligned} &\text { }\\ &\frac{d^{2} x}{d y^{2}}=0 \end{aligned}

Answers (1)

i) Integrating factor of the differential of the form \frac{d x}{d y}+P_{1} x=Q_{1} is given by e^{\int P_{1}d y}. Hence given statement is true.

 

(ii) Solution of the differential equation of the type \frac{d x}{d y^{\prime}}+P_{1} x=Q_{1}  is given by x e^{\int P_{1} d y}=\int Q_{1} e^{\int P_{1} d y} d y+C

Hence given statement is true.

iii) Correct substitution for the solution of the differential equation of the type \frac{d y}{d x} f(x, y),$ where $f(x, y) is a homogeneous function of zero degree is  y=v x. 

Hence given statement is true.

(iv) Correct substitution for the solution of the differential equation of the type \frac{d x}{d y} g(x, y)where g(x, y) is a homogeneous function of the degree zero is x=v y.

Hence given statement is true.

(V) There is no arbitrary constants in the particular solution of a differential equation. Hence given statement is Flase.

(vi) In thegiven equation x^{2}+(y-a)^{2}=$ $a^{2} the number of arbitrary constant is one. So the order order will be one. 

Hence given statement is False.

 
(vii) \frac{d y}{d x}=\left(\frac{y}{x}\right)^{1 / 3}

\frac{d y}{y\frac{1}{3}}=\left(\frac{dx}{x^\frac{1}{3}}\right)

\int \frac{d y}{y\frac{1}{3}}=\int \left(\frac{dx}{x^\frac{1}{3}}\right)

\begin{array}{l} \frac{3}{2} y^{2 / 3}=\frac{3}{2} x^{2 / 3}+C^{\prime} \\ y^{2 / 3}-x^{2 / 3}=C \end{array}

Hence the given statement is true.
(viii)  y=e^{x}(A \cos x+$$B \sin x )

\frac{d y}{d x}=e^x(-A\sin x + B \cos x) + e^x(A\sin x + B \cos x)

\frac{d y}{d x}=e^x(-A\sin x + B \cos x) + y

\frac{d^2 y}{d x^2}=e^x(-A\sin x + B \cos x) + e^x(-A\cos x - B \sin x)+\frac{dy}{dx}

\frac{d^{2} y}{d x^{2}}-2 \frac{d y}{d x}+2 y=0.

Hence the given statement is true.

ix) Given:  \frac{d y}{d x}=\frac{x+2 y}{x}

\frac{d y}{d x}-\frac{2}{x} y=1

Compare with \frac{d y}{d x}+P_{1} y=Q_{1}

Here P_{1} = \frac{-2}{x}Q_{1} = 1

I.F. = e^{\int \frac{-2}{x}dx} = e^{\log \frac{1}{x^2}} = \frac{1}{x^2}

General solution 

y. \frac{1}{x^2} = \int \frac{1}{x^2}dx

\frac{y}{x^2} = \frac{-1}{x}+c

y+x= cx^2

Hence the given statement is true.

x) Given: \frac{x d y}{d x}=y+x \tan \left(\frac{y}{x}\right)

\frac{ d y}{d x}=\frac{y}{x}+ \tan \left(\frac{y}{x}\right)

Let y =vx

\frac{ d y}{d x}=v+x\frac{dv}{dx}

v+x\frac{dv}{dx} = v+\tan v

x\frac{dv}{dx} = \tan v

\int \frac{dv}{\tan v} = \int \frac{dx}{x}

\log \sin v = \log x + \log c

\sin v = xc

\sin \frac{y}{x} = cx

Hence the given statement is true.

xi) Assume equation of a non-horizontal line in the plane

y = mx +c

\frac{dy}{dx} = m

\frac{dx}{dy} = \frac{1}{m}

 \begin{aligned} &\text { }\\ &\frac{d^{2} x}{d y^{2}}=0 \end{aligned}

Hence the given statement is true.

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