1.Singular Solutions and Extraneous Loci

Singular Solutions and Extraneous Loci

In this article “Singular Solutions and Extraneous Loci”,we will investigate for singular solution and extraneous loci of the differential equations.

Also Read This Article:- Method of Finding Singular Solution

2.Singular Solutions and Extraneous Loci Solved Examples

Investigate for singular solution and extraneous loci of the following differential equations:
Example:1. p^2(1-x^2)=1-y^2
Solution:The given equation can be written in the following form
p^2(1-x^2)=1-y^2 \cdots(1) \\p^2=\frac{1-y^2}{1-x^2}\\ p=\pm\sqrt{\frac{1-y^2}{1-x^2}}\\ p=\frac{dy}{dx}=\pm\sqrt{\frac{1-y^2}{1-x^2}}\\ \frac{dy}{\sqrt{1-y^2}}\pm\frac{dx}{\sqrt{1-x^2}}=0
Integrating both sides,we get
\sin^{-1}y\pm\sin^{-1}x=\sin^{-1}c\\ \sin^{-1} \left[ y\sqrt{1-x^2}\pm x\sqrt{1-y^2} \right]=\sin^{-1}c\\ \left[ y\sqrt{1-x^2}\pm x\sqrt{1-y^2} \right]=c \\ y\sqrt{1-x^2}=c\mp x\sqrt{1-y^2} \cdots(2)
Which is general solution
c-discriminant
Squaring (2) both sides
y^2(1-x^2)=c^2\mp2cx\sqrt{1-y^2}+x^2 \left( \sqrt{(1-y^2)} \right)^2 \\ c^2\mp2cx\sqrt{1-y^2}+x^2(1-y^2)-y^2(1-x^2)=0\\ B^2-4AC=0\\ (\mp2x\sqrt{1-y^2})^2-4 \times 1 \times \left[x^2(1-y^2)-y^2(1-x^2)\right]=0\\ 4x^2(1-y^2)-4 \left(x^2-x^2y^2-y^2+x^2y^2\right)=0 \\ 4\left[x^2-x^2y^2-x^2+y^2\right]=0\\ \Rightarrow y^2-x^2y^2=0\\ \Rightarrow y^2(1-x^2)=0 \\ \Rightarrow y^2(1-x)(1+x)=0 \cdots(3)
p-discriminant
p^2(1-x^2)+0\cdot p-(1-y^2)=0\\ \Rightarrow B^2-4AC=0\\ \Rightarrow 0^2-4(1-x^2)\left[-(1-y^2)\right]=0\\ \Rightarrow (1-x^2)(1-y^2)=0\\ \Rightarrow (1-x)(1+x)(1-y)(1+y)=0 \cdots(4)
Common factors occurring once in c-discriminant and p-discriminant (1-x)(1+x)=0 is singular solution
\Rightarrow x=\pm 1
Again y=0 which occurs twice is c-discriminant only and does not satisfy the differential equation:
Nodal Locus : y=0
Example:2. y^2(y-xp)=x^4p^2
Solution: y^2(y-xp)=x^4p^2 \cdots(1)
Put u=\frac{1}{x},\qquad v=\frac{1}{y}\\ \Rightarrow \frac{du}{dx}=-\frac{1}{x^2},\qquad \frac{dv}{dy}=-\frac{1}{y^2}\\ \Rightarrow \frac{\dfrac{du}{dx}}{\dfrac{dv}{dy}}=\frac{-\dfrac{1}{x^2}}{-\dfrac{1}{y^2}} \\ \Rightarrow\frac{dy}{dx}=\frac{y^2}{x^2}\frac{dv}{du} \\ p=\frac{y^2}{x^2}P
Putting these values in (1),we get
y^2\left(y-x\cdot\frac{y^2}{x^2}P\right)=x^4\cdot\frac{y^4}{x^4}P^2\\ \Rightarrow \frac{y^2}{x}(xy-y^2P)=y^4P^2\\ \Rightarrow \frac{u}{v^2}\left(\frac{1}{u}\cdot\frac{1}{v}-\frac{1}{v^2}P\right)=\frac{1}{v^4}P^2\\ \Rightarrow \frac{u}{v^2}\left(\frac{v-uP}{uv^2}\right)=\frac{1}{v^4}P^2\\ \Rightarrow \frac{u(v-uP)}{u\cdot v^4}=\frac{1}{v^4}P^2\\ \Rightarrow v-uP=P^2 \\ v=uP+P^2 \cdots(2)
Which is of Clairaut’s form;therefore its solution is
v=uc+c^2\\ \Rightarrow \frac{1}{y}=\frac{1}{x}c+c^2\\ \Rightarrow x=cy+c^2xy
Which is primitive:
c-discriminant
B^2-4AC=0\\ \Rightarrow y^2-4xy(-x)=0\\ \Rightarrow y^2+4x^2y=0\\ \Rightarrow y(y+4x^2)=0 \cdots(3)
p-discriminant
From (2):
P^2+uP-v=0\\ u^2-4(1)(-v)=0\\ \Rightarrow \frac{1}{x^2}+\frac{4}{y}=0\\ \Rightarrow y+4x^2=0
Common factors occurring once in (3) and (4)
y=4 x^2 is singular solution
Example:3. y = 2xp + y^2p^3
Solution:The given equation can be written in the following form
y = 2xp + y^2p^3 \cdots(1) \\ 2xp = y - y^2p^3 \\ \Rightarrow x = \frac{y}{2p} - \frac{y^2p^2}{2}
Now differentiating w.r.t. y,we get
\Rightarrow \frac{dx}{dy} = \frac{1}{2p} - \frac{y}{2p^2} \frac{dp}{dy} - yp^2 - y^2p \frac{dp}{dy} \\ \Rightarrow \frac{1}{p} = \frac{1}{2p} - \frac{y}{2p^2} \frac{dp}{dy} - yp^2 - y^2p \frac{dp}{dy} \quad \left[ \because \frac{dx}{dy} = \frac{1}{p} \right] \\ \Rightarrow \frac{1}{p} - \frac{1}{2p} + \frac{y}{2p^2} \frac{dp}{dy} + yp^2 + y^2p \frac{dp}{dy} = 0 \\ \Rightarrow \frac{1}{2p} + \frac{y}{2p^2} \frac{dp}{dy} + yp^2 + y^2p \frac{dp}{dy} = 0 \\ \Rightarrow \frac{1}{2p} \left( 1 + \frac{y}{p} \frac{dp}{dy} \right) + yp^2 \left( 1 + \frac{y}{p} \frac{dp}{dy} \right) = 0 \\ \Rightarrow \left( 1 + \frac{y}{p} \frac{dp}{dy} \right) \left( \frac{1}{2p} + yp^2 \right) = 0
Here the first factor when equated to zero will give primitive,we have
1 + \frac{y}{p} \frac{dp}{dy} = 0 \\ \Rightarrow \frac{y}{p} \frac{dp}{dy} = -1 \\ \Rightarrow \frac{dp}{p} = -\frac{dy}{y}
Integrating both sides,we get
\log p = -\log y + \log c \\ \Rightarrow p = \frac{c}{y}
Elimination of p between (1) and (2) we get
y = 2x \left( \frac{c}{y} \right) + y^2 \left( \frac{c^3}{y^3} \right) \\ \Rightarrow y = \frac{2xc}{y} + \frac{c^3}{y} \\ \Rightarrow y^2 = 2cx + c^3 \quad \cdots (3)
Differentiating w.r.t. c,we get
0=2x + 3c^2 \quad \cdots (4)
c-discriminant
Elimination c between (3) and (4)
(y^2)^2 = (2xc + c^3)^2 \\ \Rightarrow y^4 = 4x^2c^2 + c^6 + 4c^4x \\ \Rightarrow y^4 = 4x^2\left(-\frac{2x}{3}\right) + \left(-\frac{2x}{3}\right)^3 + 4\left(-\frac{2x}{3}\right)^2x \\ \Rightarrow y^4 = -\frac{8}{3}x^3 - \frac{8x^3}{27} + \frac{16x^3}{9} \\ \Rightarrow y^4 = \frac{-72x^3 - 8x^3 + 48x^3}{27} \\ \Rightarrow y^4 = \frac{-80x^3 + 48x^3}{27} \\ \Rightarrow 27y^4 = -32x^3 \\ \Rightarrow 27y^4 + 32x^3 = 0
p-discriminant
differentiating (1) w.r.t p , we get
0 = 2x + 3y^2p^2 \\ \Rightarrow p^2 = -\frac{2x}{3y^2} \quad \cdots (5)
Elimination p between (1) and (5)
y^2 = 2xp + y^2p^3 \\ y^2 = (2xp + y^2p^3)^2 \\ \Rightarrow y^2 = 4x^2p^2 + 4xy^2p^4 + y^4p^6 \\ = 4x^2\left(-\frac{2x}{3y^2}\right) + 4xy^2\left(-\frac{2x}{3y^2}\right)^2 + y^4\left(-\frac{2x}{3y^2}\right)^3 \\ = -\frac{8x^3}{3y^2} + \frac{16x^3}{9y^2} - \frac{8x^3}{27y^2} \\ = \frac{-72x^3 + 48x^3 - 8x^3}{27y^2} \\ = \frac{-32x^3}{27y^2} \\ \Rightarrow 27y^4 + 32x^3 = 0 \quad \cdots (6)
Common factors occurring once in c-discriminant and p-discriminant in (5) and (6) therefore
Singular solution: 27y^4 + 32x^3 = 0

Singular Solutions and Extraneous Loci

Example:4. 4x(x-1)(x-2)p^2 - (3x^2 - 6x + 2)^2 = 0
Solution: 4x(x-1)(x-2)p^2 = (3x^2 - 6x + 2)^2 =0 \cdots(1)\\ p^2 = \frac{(3x^2 - 6x + 2)^2}{4x(x-1)(x-2)} \\ p=\frac{dy}{dx} = \pm \frac{3x^2 - 6x + 2}{2\sqrt{x(x-1)(x-2)}} \\ \Rightarrow \frac{dy}{dx} = \pm \frac{3x^2 - 6x + 2}{2\sqrt{x^3 - 3x^2 + 2x}} \\ \Rightarrow \int dy = \pm \int \frac{3x^2 - 6x + 2}{2\sqrt{x^3 - 3x^2 + 2x}} dx
Put x^3 - 3x^2 + 2x = t \\ \Rightarrow (3x^2 - 6x + 2)dx = dt \\ \Rightarrow y = \pm \int \frac{dt}{2\sqrt{t}} \\ \Rightarrow y = \pm \sqrt{t} + c \\ \Rightarrow y = \pm \sqrt{x^3 - 3x^2 + 2x} + c \\ \Rightarrow (y-c)^2 = x^3 - 3x^2 + 2x
Integrating both sides,we get
Which is primitive of the equation,we can written as
c^2 - 2cy + y^2 - x(x-1)(x-2) = 0
c-discriminant
B^2 - 4AC = 0 \\ \Rightarrow (-2y)^2 - 4 \cdot 1 \cdot [y^2 - x(x-1)(x-2)] = 0 \\ \Rightarrow 4y^2 - 4[y^2 - x(x-1)(x-2)] = 0 \\ \Rightarrow 4[y^2 - y^2 + x(x-1)(x-2)] = 0 \\ \Rightarrow x(x-1)(x-2) = 0 \quad \cdots (2)
p-discriminant
B^2 - 4AC = 0 \\ \Rightarrow 0 - 4(x)(x-1)(x-2) [- (3x^2 - 6x + 2)^2 ]= 0 \\ \Rightarrow x(x-1)(x-2)(3x^2 - 6x + 2)^2 = 0 \quad \cdots (3)
Common factors occurring once in c-discriminant and p-discriminant in (2) and (3)
x(x-1)(x-2) = 0
Again (3x^2 - 6x + 2)^2 = 0 which is occurs twice only in p-discriminant represents tac-locus
x = \frac{6 \pm \sqrt{(-6)^2 - 4 \times 3 \times 2}}{2 \times 3} \\ = \frac{6 \pm \sqrt{36 - 24}}{6} \\ = \frac{6 \pm \sqrt{12}}{6} \\ = \frac{6 \pm 4\sqrt{3}}{6} \\ = \frac{6}{6} \pm \frac{4\sqrt{3}}{6} \\ \Rightarrow x = 1 \pm \frac{1}{\sqrt{3}}
Example:5. p^2(x^2 - a^2) - 2pxy + y^2 - b^2 = 0
Solution: p^2(x^2 - a^2) - 2pxy + y^2 - b^2 = 0 \\ y = \frac{2px \pm \sqrt{(-2px)^2 - 4 \cdot 1 \cdot [p^2(x^2 - a^2) - b^2]}}{2 \cdot 1} \\ =\frac{2px \pm \sqrt{4p^2x^2 - 4p^2x^2 + 4p^2a^2 - 4b^2}}{2} \\ = \frac{2px \pm \sqrt{4(p^2a^2 - b^2)}}{2} \\ \Rightarrow y = px \pm \sqrt{p^2a^2 - b^2}
This is clairaut’s form.Hence replacing p by c,general solution is
\Rightarrow y = cx \pm \sqrt{c^2a^2 - b^2}
p-discriminant
(a^2 - x^2)p^2 + 2xyp + (b^2 - y^2) = 0 \\ \Rightarrow B^2 - 4AC = 0 \\ \Rightarrow (2xy)^2 - 4(a^2 - x^2)(b^2 - y^2) = 0 \\ \Rightarrow 4x^2y^2 - 4(a^2b^2 - a^2y^2 - b^2x^2 + x^2y^2) = 0 \\ \Rightarrow 4[x^2y^2 - a^2b^2 + a^2y^2 + b^2x^2 - x^2y^2] = 0 \\ \Rightarrow a^2y^2 + b^2x^2 = a^2b^2 \\ \Rightarrow \frac{a^2y^2}{a^2b^2} + \frac{b^2x^2}{a^2b^2} = 1 \\ \Rightarrow \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1
Which is singular solution
Example:6. (px - y)(x - py) = 2p
Solution: (px - y)(x - py) = 2p \quad \cdots (1)
This is clairaut’s form.Hence replacing p by c,the general solution is
Put x^2 = u \text{ and } y^2 = v \\ 2xdx = du \text{ and } 2ydy = dv \\ \frac{y}{x}\left(\frac{dy}{dx}\right) = \frac{dv}{du} \Rightarrow p = \left(\frac{x}{y}\right)P \\ \Rightarrow \left(\frac{x}{y}P \cdot x - y\right)\left(x - \frac{x}{y}P \cdot y\right) = 2\frac{x}{y}P \\ \Rightarrow \frac{(x^2P - y^2)}{y} (x - xP) = \frac{2x}{y}P \\ \Rightarrow x(x^2P - y^2)(1 - P) = 2P \cdot x \\ \Rightarrow (x^2P - y^2)(1 - P) = 2P \\ \Rightarrow (uP - v)(1 - P) = 2P \\ \Rightarrow uP - uP^2 - v + vP = 2P \\ \Rightarrow uP - v = \frac{2P}{1 - P} \\ \Rightarrow v = uP - \frac{2P}{1 - P}
This is cairayt,s form.Hence replacing p byy c, the general solution is
\Rightarrow v = uc - \frac{2c}{1-c} \\ \Rightarrow y^2 = x^2c - \frac{2c}{1-c} \\ \Rightarrow y^2 = \frac{x^2c(1-c) - 2c}{1-c} \\ \Rightarrow y^2(1-c) = x^2c - x^2c^2 - 2c \\ \Rightarrow y^2 - cy^2 = x^2c - x^2c^2 - 2c \\ \Rightarrow x^2c^2 + (-x^2 - y^2 + 2)c + y^2 = 0 \\ \Rightarrow c^2x^2 - (x^2 + y^2 - 2)c + y^2 = 0
c-discriminant
B^2 - 4AC = 0 \\ \Rightarrow [-(x^2 + y^2 - 2)]^2 - 4 \cdot x^2 \cdot y^2 = 0 \\ \Rightarrow (x^2 + y^2 - 2)^2 - 4x^2y^2 = 0 \\ \Rightarrow (x^2 + y^2 - 2)^2 - (2xy)^2 = 0 \\ \Rightarrow (x^2 + y^2 - 2 - 2xy)(x^2 + y^2 - 2 + 2xy) = 0 \\ \Rightarrow [(x-y)^2 - (\sqrt{2})^2] [(x+y)^2 - (\sqrt{2})^2] = 0 \\ \Rightarrow (x-y-\sqrt{2})(x-y+\sqrt{2})(x+y+\sqrt{2})(x+y-\sqrt{2}) = 0 \quad \cdots (1)
p-discriminant
(px - y)(x - py) = 2p \\ \Rightarrow px^2 - p^2xy - xy + py^2 = 2p \\ \Rightarrow p^2xy - p(x^2 + y^2 - 2) + xy = 0 \\ B^2 - 4AC = 0 \\ \Rightarrow [-(x^2 + y^2 - 2)]^2 - 4 \cdot xy \cdot xy = 0 \\ \Rightarrow (x^2 + y^2 - 2)^2 - 4x^2y^2 = 0 \\ \Rightarrow (x^2 + y^2 - 2)^2 - (2xy)^2 = 0 \\ \Rightarrow (x^2 + y^2 - 2 - 2xy)(x^2 + y^2 - 2 + 2xy) = 0 \\ \Rightarrow [(x-y)^2 - (\sqrt{2})^2] [(x+y)^2 - (\sqrt{2})^2] = 0 \\ \Rightarrow (x-y-\sqrt{2})(x-y+\sqrt{2})(x+y-\sqrt{2})(x+y+\sqrt{2}) = 0 \quad \cdots (2)
Common factors occurring once in c-discriminant and p-discriminant in (1) and (2)
(x-y-\sqrt{2})(x-y+\sqrt{2})(x+y-\sqrt{2})(x+y+\sqrt{2}) = 0
is singular solution
Example:7. p^3 - 4xyp + 8y^2 = 0
Solution: p^3 - 4xyp + 8y^2 = 0
The given equation can be written in the following form
4xyp = p^3 + 8y^2 \\ \Rightarrow x = \frac{p^3}{4yp} + \frac{8y^2}{4yp} \Rightarrow x = \frac{p^2}{4y} + \frac{2y}{p}
Differentiating both sides w.r.t. y,we get
\frac{dx}{dy} = -\frac{p^2}{4y^2} + \frac{p}{2y}\frac{dp}{dy} + \frac{2}{p} - \frac{2y}{p^2}\frac{dp}{dy} \\ \Rightarrow \frac{1}{p} = -\frac{p^2}{4y^2} + \frac{p}{2y}\frac{dp}{dy} + \frac{2}{p} - \frac{2y}{p^2}\frac{dp}{dy} \quad \left(\text{since } \frac{dx}{dy} = \frac{1}{p}\right) \\ \Rightarrow \frac{1}{p} - \frac{2}{p} + \frac{p^2}{4y^2} - \frac{p}{2y}\frac{dp}{dy} + \frac{2y}{p^2}\frac{dp}{dy} = 0 \\ \Rightarrow -\frac{1}{p} - \frac{p}{2y}\frac{dp}{dy} + \frac{p^2}{4y^2} + \frac{2y}{p^2}\frac{dp}{dy} = 0 \\ \Rightarrow -1\left(\frac{1}{p} + \frac{p}{2y}\frac{dp}{dy}\right) + \frac{p^2}{4y^2}\left(\frac{1}{p} + \frac{p}{2y}\frac{dp}{dy}\right) = 0 \\ \Rightarrow -\frac{1}{p} + \frac{2y}{p^2}\frac{dp}{dy} - \frac{p}{2y}\frac{dp}{dy} + \frac{p^2}{4y^2} = 0 \\ \Rightarrow -\frac{1}{p}\left(1 - \frac{2y}{p}\frac{dp}{dy}\right) + \frac{p^2}{4y^2}\left(1 - \frac{2y}{p}\frac{dp}{dy}\right) = 0 \\ \Rightarrow \left(1 - \frac{2y}{p}\frac{dp}{dy}\right) \left(-\frac{1}{p} + \frac{p^2}{4y^2}\right) = 0
Here the first factor when equated to zero will give primitive we have
1 - \frac{2y}{p}\frac{dp}{dy} = 0 \\ \Rightarrow \frac{2y}{p}\frac{dp}{dy} = 1 \\ \Rightarrow \frac{dy}{y} = \frac{2}{p} dp
Integrate : \int \frac{2}{p} dp = \int \frac{dy}{y} \\ \Rightarrow 2\log p = \log y + \log c_1 \\ \Rightarrow p^2 = c_1y \quad \cdots (2)
Elemininating p from (1) and (2),we get
(c_1y)^{3/2} - 4xy(c_1y)^{1/2} + 8y^2 = 0 \\ \Rightarrow y^{3/2} [c_1^{3/2} \cdot y - 4c_1^{1/2}x + 8y^{1/2}] = 0 \\ \Rightarrow c_1^{3/2}y - 4c_1^{1/2}x + 8y^{1/2} = 0 \\ \Rightarrow 8y^{1/2} = 4c_1^{1/2}x - c_1^{3/2}y \\ \Rightarrow y^{1/2} = \frac{c_1^{1/2}}{2}x - \frac{c_1^{3/2}}{8} \\ y^{1/2} = \frac{c_1^{1/2}}{2}(x - \frac{c_1}{4})
c-discriminant
put \frac{c_1^{1/2}}{2} = \sqrt{C} \\ y^{1/2} = \sqrt{C}(x-C) \\ \Rightarrow y = C(x-C)^2 \quad \cdots (3)
Differentiating w.r.t. c,we get
0 = (x-C)^2 + C \cdot 2(x-C)(-1) \\ \Rightarrow (x-C) [x-C-2C] = 0 \\ \Rightarrow (x-C)(x-3C) = 0 \quad \text{put } C = \frac{x}{3} \\ y = \frac{x}{3}(x-\frac{x}{3})^2 \Rightarrow y = \frac{x}{3} \cdot \frac{4x^2}{9} \\ \Rightarrow y(27y - 4x^3) = 0 \quad \cdots (4)
p-discriminant
Differentiating (1) w.r.t. p,we get
3p^2 - 4xy = 0 \Rightarrow p^2 = \frac{4}{3}xy \quad \cdots (5)
Eliminating p from (1) and (5)
(p^3 - 4xyp)^2 = (-8y^2)^2 \\ \Rightarrow p^6 + 16x^2y^2p^2 - 8xyp^4 = 64y^4 \\ \Rightarrow (\frac{4}{3}xy)^3 + 16x^2y^2(\frac{4}{3}xy) - 8xy(\frac{4}{3}xy)^2 = 64y^4 \\ \Rightarrow \frac{64}{27}x^3y^3 + \frac{64}{3}x^3y^3 - \frac{128}{9}x^3y^3 = 64y^4 \\ \Rightarrow \frac{64}{27}x^3y^3 - \frac{128}{9}x^3y^3 + \frac{64}{3}x^3y^3 -64y^4=0 \\ \Rightarrow \frac{64x^3y^3-384x^3y^3}{27} + \frac{64}{3}x^3y^3 -64y^4=0 \\ \Rightarrow -\frac{320x^3y^3}{27} + \frac{64}{3}x^3y^3 -64y^4=0 \\ \Rightarrow \frac{-320x^3y^3 + 576x^3y^3}{27} - 64y^4 = 0 \\ \Rightarrow \frac{256x^3y^3}{27} - 64y^4 = 0 \\ \Rightarrow \frac{64y^3}{27}(4x^3 - 27y) = 0 \\ y^3(27y-4x^3)=0\cdots(6)
Common factors occurring once in c-discriminant and p-discriminant in (5) and (6)
\Rightarrow y(27y-4x^3)=0
is singular solution
With the above illustrations,one can understand the Singular Solutions and Extraneous Loci.

Also Read This Article:- Singular Solution of Differential Equation

3.Practice Questions of Singular Solutions and Extraneous Loci for Students

Singular Solutions and Extraneous Loci

Investigate for singular solution and extraneous loci of the following differential equations:
(1.)y^4=4y(xp-2y)^2
(2.)p^2+2px^3-4x^2y=0
Answers:(1.)Singular solution x^4-16y=0 ,y=0
(2.)Primitive y-cx^2-c^2=0 ,Singular Solution x^2+xy=0, Tac-Locus x=0
By solving the above questions,you can understand the Singular Solutions and Extraneous Loci well because the concept is well understood when you solve it practically
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Also Read This Article:- Singular solution

4.Frequently Asked Questions Related to Singular Solutions and Extraneous Loci

\begin{array}{|c|} \hline \text{**छात्र-छात्राओं से आज का प्रश्न**} \\ \text{एक घनाभ का पृष्ठीय क्षेत्रफल 22 वर्गसेमी है} \\ \text{ तथा उसकी सभी कोरों की लम्बाईयों का योग 24 सेमी है।} \\ \text{इस घनाभ के प्रत्येक विकर्ण की लम्बाई (सेमी में) होगी?} \\ \text{दिनांक 29.06.2026 के प्रश्न का उत्तर:10 kg} \\ \text{Let weight of Shelf=x,Weight of Book=y} \\ x+y=25 \cdots(1) \\ x+\frac{1}{3}y=15 \cdots(2) \\ \text{Multiply (1) by } \frac{1}{8} \\ \frac{1}{3} x+\frac{1}{3} y=\frac{25}{3} \cdots(5) \\ \text{Subtract (2) from (3),we get} \\ \frac{2}{3}x=\frac{20}{3} \\ \Rightarrow x=10 \text{kg} \\ \text{**Today's Question to Students**} \\ \text{*"The surface area of a cuboid is 22 square} \\ \text{ cm and the sum of the lengths of all its} \\ \text{ cores is 24 cm.What is the length (in cm)} \\ \text{ of each diagonal of this cuboid?" *} \\ \text{Answer to Question Dated 29.06.2026:10 kg.}  \\  \text{Let weight of Shelf=x,Weight of Book=y} \\ x+y=25 \cdots(1) \\ x+\frac{1}{3}y=15 \cdots(2) \\ \text{Multiply (1) by } \frac{1}{8} \\ \frac{1}{3} x+\frac{1}{3} y=\frac{25}{3} \cdots(5) \\ \text{Subtract (2) from (3),we get} \\ \frac{2}{3}x=\frac{20}{3} \\ \Rightarrow x=10 \text{kg} \\ \hline  \end{array}
This article has been prepared by **Satyam Coaching Centre** on the **Satyam Mathematics** blog.”*