(i) The degree of the differential equation $\frac{{{d^2}y}}{{d{x^2}}} + {e^{dy/dx}} = 0$ is……………….

Given differential equation is

$\frac{{{d^2}y}}{{d{x^2}}} + {e^{\frac{{dy}}{{dx}}}} = 0$

Degree of this equation is not defined.

(ii) The degree of the differential equation $\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} = x$ is……………….

Solution

Given differential equation is $\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} = x$

So, degree of this equation is two.

(iii) The number of arbitrary constants in the general Solution of a differential equation of order three is .................

Solution

There are three arbitrary constants.

(iv) $\frac{{dy}}{{dx}} + \frac{y}{{x\log x}} = \frac{1}{x}$ is an equation of the type …………….

Solution

Given differential equation is

$\frac{{dy}}{{dx}} + \frac{y}{{x\log x}} = \frac{1}{x}$

The equation is of the type

$\frac{{dy}}{{dx}} + Py = Q$

(v) General Solution of the differential equation of the type is given by ……………

Solution

Given differential equation is
$\frac{{dx}}{{dy}} + {P_1}x = {Q_1}$
The general Solution is

$x \cdot {\rm{IF}} = \int Q ({\rm{IF}})dy + C{\rm{ i}}{\rm{.e}}{\rm{., }}x{e^{\int P dy}} = \int Q \left\{ {{e^{\int P y}}} \right\}dy + C$

(vi) The Solution of the differential

equation $\frac{{xdy}}{{dx}} + 2y = {x^2}$
is .................

Solution

Given differential equation is

$x\frac{{dy}}{{dx}} + 2y = {x^2} \Rightarrow \frac{{dy}}{{dx}} + \frac{{2y}}{x} = x$

This equation of the form
$\frac{{dy}}{{dx}} + Py = Q$.

$therefore ~ {\rm{IF}} = {e^{\int {\frac{2}{x}} dx}} = {e^{2\log x}} = {x^2}$

The general Solution is
$y{x^2} = \int x \cdot {x^2}dx + C$

$\Rightarrow$ $y{x^2} = \frac{{{x^4}}}{4} + C$

$\Rightarrow$ $y = \frac{{{x^2}}}{4} + C{x^{ - 2}}$

(vii) The Solution of $\left( {1 + {x^2}} \right)\frac{{dy}}{{dx}} + 2xy - 4{x^2} = 0$ is ................

Solution

Given differential equation is

$\left( {1 + {x^2}} \right)\frac{{dy}}{{dx}} + 2xy - 4{x^2} = 0$

$\Rightarrow$ $\frac{{dy}}{{dx}} + \frac{{2xy}}{{1 + {x^2}}} - \frac{{4{x^2}}}{{1 + {x^2}}} = 0$

$\Rightarrow$ $\frac{{dy}}{{dx}} + \frac{{2x}}{{1 + {x^2}}}y = \frac{{4{x^2}}}{{1 + {x^2}}}$

$$\therefore $$ ${\rm{IF}} = {e^{\int {\frac{{2x}}{{1 + {x^2}}}} dx}}$

Put $1 + {x^2} = t \Rightarrow 2xdx = dt$

$\therefore {\rm{IF}} = {e^{\int {\frac{{{\rm{dt}}}}{t}} }} = {e^{\log t}} = {e^{\log \left( {1 + {x^2}} \right)}} = 1 + {x^2}$

The general Solution is

$y \cdot \left( {1 + {x^2}} \right) = \int {\left( {1 + {x^2}} \right)} \frac{{4{x^2}}}{{\left( {1 + {x^2}} \right)}}dx + C$

$\Rightarrow$ $\left( {1 + {x^2}} \right)y = \int 4 {x^2}dx + C$
$\Rightarrow$ $\quad \left( {1 + {x^2}} \right)y = 4\frac{{{x^3}}}{3} + C$

$\Rightarrow$ $y = \frac{{4{x^3}}}{{3\left( {1 + {x^2}} \right)}} + C{\left( {1 + {x^2}} \right)^{ - 1}}$

(viii) The Solution of the differential equation $ydx + (x + xy)dy = 0$ is…….........

Solution

Given differential equation is

$\Rightarrow$ $ydx + (x + xy)dy = 0$

$\Rightarrow$ $ydx + x(1 + y)dy = 0$

$\Rightarrow$ $\frac{{dx}}{{ - x}} = \left( {\frac{{1 + y}}{y}} \right)dy$

$\Rightarrow$ $\int {\frac{1}{x}} dx = - \int {\left( {\frac{1}{y} + 1} \right)} dy\quad$

[on integrating]
$\Rightarrow$ $\log (x) = - \log (y) - y + \log A$

$\log (x) + \log (y) + y = \log A$

$\log (xy) + y = \log A$

$\Rightarrow$ $\log xy + \log {e^y} = \log A$

$\Rightarrow$ $xy{e^y} = A$

$\Rightarrow$ $xy = A{e^{ - y}}$

(ix) General Solution of $\frac{{dy}}{{dx}} + y = \sin x$ is ………………

Solution

Given differential equation is

$\frac{{dy}}{{dx}} + y = \sin x$

${\rm{IF}} = \int {{e^{1dx}}} = {{\rm{e}}^x}$

The general Solution is

$y \cdot {e^x} = \int {{e^x}} \sin xdx + C$
…..(i)
Let $I = \int {{e^x}} \sin xdx$

$I = \sin x{e^x} - \int {\cos } x{e^x}dx$

$= \sin x{e^x} - \cos x{e^x} + \int {( - \sin x)} {e^x}dx$

$2I = {e^x}(\sin x - \cos x)$

$I = \frac{1}{2}{e^x}(\sin x - \cos x)$

From Eq. (i),

$y \cdot {e^x} = \frac{x}{2}(\sin x - \cos x) + C$

$\Rightarrow$ $y = \frac{1}{2}(\sin x - \cos x) + C \cdot {e^{ - x}}$

(x) The Solution of differential equation cot $ydx = xdy$ is ………………

Solution

Given differential equation is

$\cot ydx = xdy$

$\Rightarrow$ $\frac{1}{x}dx = \tan ydy$

On integrating both sides,

we get
$\Rightarrow$ $\int {\frac{1}{x}} dx = \int {\tan } ydy$

$\Rightarrow$ $\log (x) = \log (\sec y) + \log C$

$\Rightarrow$ $\log \left( {\frac{x}{{\sec y}}} \right) = \log C$

$\Rightarrow$ $\frac{x}{{\sec y}} = C$

$\Rightarrow$ $x = C$ sec $y$

(xi) The integrating factor of
$\frac{{dy}}{{dx}} + y = \frac{{1 + y}}{x}$ is………………….

Solution

Given differential equation is

$\frac{{dy}}{{dx}} + y = \frac{{1 + y}}{x}$

$\frac{{dy}}{{dx}} + y = \frac{1}{x} + \frac{y}{x}$

$\Rightarrow$ $\frac{{dy}}{{dx}} + y\left( {1 - \frac{1}{x}} \right) = \frac{1}{x}$

$\therefore$ ${\rm{IF}} = {e^{\int {\left( {1 - \frac{1}{x}} \right)} dx}}$

$= {e^{x - \log x}}$

$= {e^x} \cdot {e^{ - \log x}} = \frac{{{e^x}}}{x}$

TRUE FALSE