Exercise 10.3 - Chapter 10 Ordinary Differential Equations 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $10.3$
Question 1.
Find the differential equation of the family of
(i) all non-vertical lines in a plane
Solution:
Equation of family of all non-vertical lines is
$\mathrm{y}=\mathrm{mx}+\mathrm{c}(\mathrm{m} \neq 0)$
Differentiate with respect to ' $x$ '
$
\frac{d y}{d x}=m \quad \text { (' } m \text { ' and ' } c \text { ' are arbitrary constants) }
$
Again, $\frac{d^2 y}{d x^2}=0$.
(ii) all non-horizontal lines in a plane.
Solution:
Equation of family of all non-horizontal lines is
$
\begin{aligned}
y & =m x+c \\
\frac{d y}{d x} & =m \Rightarrow \frac{d x}{d y}=\frac{1}{m} \\
\frac{d^2 x}{d y^2} & =0 .
\end{aligned}
$
Question 2.
Form the differential equation of all straight lines touching the circle $x^2+y^2=r^2$
Solution:
Equation of circle $x^2+y^2=r^2$ of the line $y=m x+c$ is to be a tangent to the circle, then the equation of the tangent is
$
y=m x \pm r \sqrt{1+m^2}
$

Differentiating with respect to $\mathrm{V}$ dy
$
\frac{d y}{d x}=m
$
Substitute in (1)
$
\begin{aligned}
y & =x\left(\frac{d y}{d x}\right) \pm r \sqrt{1+\left(\frac{d y}{d x}\right)^2} \\
y-x\left(\frac{d y}{d x}\right) & =\pm r \sqrt{1+\left(\frac{d y}{d x}\right)^2}
\end{aligned}
$
Squaring on both sides
$
\begin{aligned}
{\left[y-x\left(\frac{d y}{d x}\right)\right]^2 } & =r^2\left[1+\left(\frac{d y}{d x}\right)^2\right] \\
{\left[x\left(\frac{d y}{d x}\right)-y\right]^2 } & =r^2\left[1+\left(\frac{d y}{d x}\right)^2\right]
\end{aligned}
$

Question 3.
Find the differential equation of the family of circles passing through the origin and having their centres on the $\mathrm{x}$-axis.
Solution:
All circles passing through the origin and having their centre on the $\mathrm{x}$-axis say at $(\mathrm{a}, 0)$ will have radius 'a' units.
$\therefore$ Equation of circle is $\left(\mathrm{x}-\mathrm{a}^2\right)+\mathrm{y}^2=\mathrm{a}^2 \ldots$..(1) $[\because$ ' $\mathrm{a}$ ' arbitrary constant $]$ Differentiate with respect to ' $x$ '
$
\begin{aligned}
2(x-a)+2 y \frac{d y}{d x} & =0 \quad(\div 2) \\
(x-a)+y \frac{d y}{d x} & =0 \Rightarrow a=x+y \frac{d y}{d x}
\end{aligned}
$
Substituting in (1)
$
\begin{aligned}
y^2\left(\frac{d y}{d x}\right)^2+y^2 & =\left(x+y \frac{d y}{d x}\right)^2 \\
y^2\left(\frac{d y}{d x}\right)^2+y^2 & =x^2+2 x y \frac{d y}{d x}+y^2\left(\frac{d y}{d x}\right)^2 \\
x^2+2 x y \frac{d y}{d x}-y^2 & =0 .
\end{aligned}
$

Question 4.
Find the differential equation of the family of all the parabolas with latus rectum $4 \mathrm{a}$ and whose axes are parallel to the $\mathrm{x}$-axis.
Solution:
Equation of all parabolas whose axis is parallel to $\mathrm{X}$ - axis is $(\mathrm{y}-\mathrm{k})^2=4 \mathrm{a}(\mathrm{x}-\mathrm{h})$
Where $(h, k)$ is the vertex
$
(y-k)^2=4 a(x-h)
$
(1) $[\because h, k$ are arbitrary constants $]$
Differentiate with respect to ' $x$ '

$
\begin{aligned}
2(y-k) \frac{d y}{d x}=4 a \Rightarrow(y-k) \frac{d y}{d x} & =2 a \\
\therefore y-k & =\frac{2 a}{(d y / d x)}
\end{aligned}
$
Now differentiating (2) with respect to ' $x$ '
$
\begin{aligned}
(y-k) \frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)\left(\frac{d y}{d x}\right) & =0 \\
\left.\frac{2 a}{(d y / d x}\right) \cdot \frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^2 & =0 \quad\left[\because y-k=\frac{2 a}{(d y / d x)}\right] \\
2 a \cdot\left(\frac{d^2 y}{d x^2}\right)+\left(\frac{d y}{d x}\right)^3 & =0 \\
\text { i.e., } 2 a y^{\prime \prime}+y^{33} & =0
\end{aligned}
$

Question 5.
Find the differential equation of the family of parabolas with vertex at $(0,-1)$ and having axis along the $\mathrm{y}$ - axis.
Solution:
Given, vertex $(0,-1)$ and axis along $y$-axis
Equation of Parabola, $(x+1)^2=-4$ ay
(1) $[\because \mathrm{a}$ is the perameter $]$
Differentiate with respect to ' $x$ '
$
\begin{aligned}
2(x+1) & =-4 a y^{\prime} \\
\frac{2(x+1)}{y^{\prime}} & =-4 a
\end{aligned}
$
Substitute in (1) $\Rightarrow(x+1)^2=2\left(\frac{x+1}{y^{\prime}}\right) y$
$
\begin{aligned}
(x+1) & =\frac{2 y}{y^{\prime}} \\
(1+x) y^{\prime}-2 y & =0
\end{aligned}
$

Question 6.
Find the differential equations of the family of all the ellipses having foci on the $y$-axis and centre at the origin.
Solution:
Equations of the family of all the Ellipses having foci on the $y$ - axis and centre at the origin is
$
\frac{x^2}{b^2}+\frac{y^2}{a^2}=1
$
$\ldots$ (1) $\quad[\because a, b$ are parameters $]$
Differentiate with respect to ' $x$ '
$
\begin{aligned}
\frac{2 x}{b^2}+\frac{2 y y^{\prime}}{a^2} & =0 \\
(\div 2) & \Rightarrow \frac{x}{b^2}+\frac{y y^{\prime}}{a^2}=0
\end{aligned}
$
Differentiate with respect to ' $x$ '
$
\begin{aligned}
\frac{1}{b^2}+\frac{1}{a^2}\left(y y^{\prime \prime}+y^{\prime 2}\right) & =0 \\
\Rightarrow \quad \frac{1}{b^2} & =\frac{-1}{a^2}\left(y y^{\prime \prime}+y^{\prime 2}\right)
\end{aligned}
$
Substitute in (2) $\Rightarrow \frac{-1}{a^2} x\left(y y^{\prime \prime}+y^{\prime 2}\right)+\frac{y y^{\prime}}{a^2}=0$
$
\begin{array}{r}
-x y y^{\prime \prime}-x y^{\prime 2}+y y^{\prime}=0 \\
x y^{\prime 2}+x y y^{\prime \prime}-y y^{\prime}=0
\end{array}
$
Question 7.
Find the differential equation corresponding to the family of curves represented by the equation $y=$ $\mathrm{Ae}^{8 \mathrm{x}}+\mathrm{Be}^{-8 \mathrm{x}}$, where $\mathrm{A}$ and $\mathrm{B}$ are arbitrary constants.

Solution:
$y=\mathrm{A} e^{8 x}+\mathrm{B} e^{-8 x} \quad$ where $\mathrm{A}$ and $\mathrm{B}$ are arbitrary constants.
Differentiate with respect to ' $x$ '
$
\frac{d y}{d x}=8 \mathrm{~A} e^{8 x}-8 \mathrm{~B} e^{-8 x}
$
Differentiate again with respect to ' $x$ '
$
\begin{aligned}
\frac{d^2 y}{d x^2} & =64 \mathrm{~A} e^{8 x}+64 \mathrm{~B} e^{-8 x}=64\left(\mathrm{~A} e^{8 x}+\mathrm{B} e^{-8 x}\right) \\
\frac{d^2 y}{d x^2} & =64 y
\end{aligned}
$
Question 8.
Find the differential equation of the curve represented by $x y=a e^x+b e^{-x}+x^2$.
Solution:
$
\begin{aligned}
& x y=a e^x+b e^{-x}+x^2 \\
& x y-x^2=a e^x+b e^{-x}
\end{aligned}
$
(1) $\left[\because a^{\prime}, b^{\prime}\right.$ are arbitrary constants $]$
Differentiate with respect to ' $x$ '
$
\mathrm{xy}^{\prime}+\mathrm{y}-2 \mathrm{x}=a \mathrm{e}^{\mathrm{x}}-b \mathrm{e}^{-\mathrm{x}}
$
Again, Differentiate with respect to ' $x$ '
$
\begin{aligned}
x y^{\prime \prime}+y^{\prime}+y^{\prime}-2 & =a e^x+b e^{-x} \\
x y^{\prime \prime}+2 y^{\prime}-2 & =a e^x+b e^{-x} \\
x y^{\prime \prime}+2 y^{\prime}-2 & =x y-x^2 \quad[\because \text { from (1)] } \\
x y^{\prime \prime}+2 y^{\prime}+x^2-x y-2 & =0
\end{aligned}
$