\(\begin{array}{c|c} \hline \text { Corner points } & \text { Value of Z } \\ \hline A(40,15) & 285 \\ B(15,20) & 150(\mathrm{~min}) \\ C(2,7 \mathrm{f}) & 228 \\ \hline \end{array}\) 
= Minimum Z = 150, when x = 15, y = 20.

Given equation is \(\frac{d}{d x}\left\{\left(\frac{d y}{d x}\right)^{3}\right\}=0\) 
\(\Rightarrow 3\left(\frac{d y}{d x}\right)^{2} \frac{d^{2} y}{d x^{2}}=b\) 
Clearly, the highest order derivative occurring in the differential equation is \(d^{2} y / d x^{2}\).
So, its order is 2. Also, it is a polynomial equation in derivatives and highest power raised to \(d^{2} y / d x^{2}\).1, so its degree is 1.
Hence, the sum of the order and degree of the above differential equation is 2 + 1 = 3.

\(\pi{a}{b}sq.units\)

\(\frac{d y}{d x} =-2 A \sin 2 x+2 B \cos 2 x \)
\(\text { and } \frac{d^{2} y}{d x^{2}} =-4 A \cos 2 x-4 B \sin 2 x=-4 y \)
\(\Rightarrow \frac{d^{2} y}{d x^{2}}+4 y =0 \)

\(\Rightarrow\) e^y . x = e^y + y + c is the required solution.

Given equations of curves are
\(y=x^{2}+2 \text { or } x^{2}=y-2\)   ..(i)
y = x ...(ii)
x = 0 ...(iii)
and x = 3 ..(iv)
Equation (i) represents a parabola of the form \(x^{2}=4 ay\) with vertex at (0, 2) and axis along positive direction of y -axis
Equation (ii) represent a line which intersect the coordinate axes at (0, 0).
Equation (iii) and (iv) represent the lines perpendicular to the x-axis.
The graph of given curves is shown below

It is clear from the figure that the required region is OABCO.
Note that the line x = 3 intersects y= x at(3,3) and \(y=x^{2}+2 \text { at }(3,11)\) .
\(\therefore\) Required area = Area of shaded portion OABCO
\( =\int_{0}^{3}\left[y_{2} \text { (parabola) }-y_{1} \text { (line) }\right] d x \)
\(=\int_{0}^{3}\left(x^{2}+2-x\right) d x=\left[\frac{x^{3}}{3}-\frac{x^{2}}{2}+2 x\right]_{0}^{3} \)
\(=\frac{27}{3}-\frac{9}{2}+6=9+6-\frac{9}{2}=15-\frac{9}{2} \)
\(=\frac{30-9}{2}=\frac{21}{2} \mathrm{sq} \text { units } \) 
Hence, the required area is \(\frac{21}{2} \text { sq units }\) .

Given region \( \left\{(x, y): x^{2}+y^{2} \leq 4, x+y \geq 2\right\}\)
 Corresponding equations are \( x^{2}+y^{2}=4 \text { , }\)
 Centre : (0,0) and radius : 2  and line x + y = 2
On plotting the inequations \( x^{2}+y^{2} \leq 4 \text { and } x+y \geq 2 \text { , }\)
 we have to find the shaded portion as required area.
\(\text {Area } =\int_{0}^{2}\left\{\sqrt{4-x^{2}}-(2-x)\right\} d x \)
\(=\left[\frac{x}{2} \sqrt{4-x^{2}}+\frac{4}{2} \sin ^{-1} \frac{x}{2}-2 x+\frac{x^{2}}{2}\right]_{0}^{2} \)
\(=\left(\frac{2}{2} \sqrt{0}+2 \sin ^{-1} 1-4+2\right)-0=(\pi-2) \mathrm{sq} \text { units } \)

We have, \(\frac{d y}{d x}-y=\sin x\), which is a linear differential equation of the form
\(\frac{d y}{d x}+P y=Q\), here P = -1 and Q = sin x
\(\therefore \quad \mathrm{IF}=e^{\int P d x}=e^{\int(-1) d x}=e^{-x}\)
Now, the general solution of given differential equation is given by \(\begin{aligned} y \cdot(\mathrm{IF}) & =\int(\mathrm{IF}) \cdot Q d x+C \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad y \cdot e^{-x} & =\int e^{-x} \sin x d x+C \end{aligned}\)     ...(i)
Let \(I=\int_{II}^{e^{-x}} \)\(\underset{1}{sin}\) x.dx                                   ...(ii)
By using the method of integration by parts, we get
\(\begin{aligned} I & =\sin x \frac{e^{-x}}{(-1)}-\int \cos x \frac{e^{-x}}{(-1)} d x \end{aligned}\)
= -sin x e^-x + \(\int e_{\mathrm{II}}^{-x}\) \(\underset{I}{cos}\) x dx
Again, by using integration by parts, we get
\(\begin{aligned} I & =-\sin x e^{-x}+\cos x \frac{e^{-x}}{(-1)}-\int(-\sin x) \frac{e^{-x}}{(-1)} d x \end{aligned}\)
\(\begin{aligned} =-\sin x e^{-x}-\cos x e^{-x}-\int e^{-x} \sin x d x \end{aligned}\)
= -sin x e^-x - cos x e^-x - I    [from Eq. (ii)]
\(\Rightarrow\)  2I = -e^-x (sin x + cos x)
\(\Rightarrow \quad I=-\frac{e^{-x}}{2}(\sin x+\cos x)\)
Then, from Eq. (1), we get
\(\begin{aligned} y \cdot e^{-x} & =-\frac{e^{-x}}{2}(\sin x+\cos x)+C \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad y & =-\frac{1}{2}(\sin x+\cos x)+C e^x \end{aligned}\)

\(\frac{15}{2} \text { sq units }\)

Given, x² ≤ y..(i)
and y ≤ |x|...(ii)
Clearly, curve (i) is parabola directed upward with vertex at (0, 0) and symmetrical about y-axis.
Also \(y=|x|=\begin{cases} x\quad if\quad x\ge 0 \\ -x\quad ,if\quad x<\quad 0 \end{cases}\)
The lines y = x and y = - x both passes through origin and have slope of + 1& - 1 respectively.

⇒ Required Area = 2x Standard Area on a side
= \(-\left[ \left( \frac { 1 }{ 2 } -1 \right) -\left( \frac { 16 }{ 2 } -4 \right) \right] \)
= \(2{ \left[ \frac { { x }^{ 2 } }{ 2 } -\frac { { x }^{ 3 } }{ 2 } \right] }_{ 0 }^{ 1 }\)
= \(2\left[ \frac { 1 }{ 2 } -\frac { 1 }{ 3 } \right] =2\times \frac { 1 }{ 6 } \)
= \(\frac { 1 }{ 3 } \) sq.units

Let x and y be nut packages and bolt packages produced each day, respectively.
\(\therefore\)  LPP is Maximize Z = 24x + 18y
Subject to 2x + 3y \(\le \) 10
3x + 2y \(\le \) 10
x, y \(\ge \) 0

Vertices are
\(A\left( 0,\frac { 10 }{ 3 } \right) ,B(2,2)\quad and\quad C\left( \frac { 10 }{ 3 } ,0 \right) \)

Hence, 2 nuts & 2 bolts to be produced to get max. profit of Rs. 84.

We have, (x² + xy) dy = (x² + y²)dx
\(\Rightarrow \quad \frac{d y}{d x}=\left(\frac{x^2+y^2}{x^2+x y}\right)\)     ...(i)
This is a homogeneous differential equation.
On putting \(y=v x \Rightarrow \frac{d y}{d x}=v \cdot 1+x \frac{d v}{d x}\) in Eq. (i),
we get
\(\begin{aligned} v+x \frac{d v}{d x} & =\left(\frac{x^2+v^2 x^2}{x^2+x \cdot x v}\right) \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad x \frac{d v}{d x} & =\frac{1+v^2}{1+v}-v=\frac{1+v^2-v-v^2}{1+v}=\frac{1-v}{1+v} \end{aligned}\)
\(\therefore\left(\frac{1+v}{1-v}\right) d v=\frac{1}{x} d x\)
On integrating both sides, we get
\(\begin{aligned} & \int\left(\frac{1+v}{1-v}\right) d v=\int \frac{1}{x} d x \end{aligned}\)
\(\begin{aligned} & \Rightarrow \quad \int\left[-1+\frac{2}{1-v}\right] d v=\log |x|+\log C \end{aligned}\)
\(\begin{aligned} & \Rightarrow \quad-v-2 \log (1-v)=\log |x|+\log C \end{aligned}\)
\(\begin{aligned} & \Rightarrow-v=2 \log (1-v)+\log |x|+\log C \end{aligned}\)
\(\begin{aligned} & \Rightarrow \quad-v=\log (1-v)^2+\log \{C|x|\} \end{aligned}\)
                                  [\(\because\) log m + log n = log mn]
\(\Rightarrow\)    -v = log{C |x| (1 - v)²}
\(\Rightarrow\)    C|x|(1 - v)² = e^-v
\(\Rightarrow C|x|\left(1-\frac{y}{x}\right)^2=e^{-y / x} \quad\left[\because v=\frac{y}{x}\right]\)   ...(ii)
On puting x = 1 and y = 0 in Eq. (ii), we get
C.1(1 - 0) = e⁰ \(\Rightarrow\) C = 1
Thus, the required solution is
\(\begin{aligned} |x|\left(1-\frac{y}{x}\right)^2 & =e^{-y / x} \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad(x-y)^2 & =|x| e^{-y / x} \end{aligned}\)
which is the required particular solutions.

= y

The given equation can be rewritten as
\(x \sqrt{1-y^{2}} d x=-y \sqrt{1-x^{2}} d y \Rightarrow \frac{d y}{d x}=\frac{-x \sqrt{1-y^{2}}}{y \sqrt{1-x^{2}}} \)
Since, the highest order derivative is \( \frac{d y}{d x}.\)is 1. So, its degree is 1.