Exercise 4.1 - Chapter 4 Differential Equations 12th Maths Guide Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Chapter 4 Differential Equations
$\operatorname{Ex} 4.1$
Question 1.
Find the order and degree of the following differential equations.
(i) $\frac{d y}{d x}+2 y=x^3$
(ii) $\frac{d^3 y}{d x^3}+3\left(\frac{d y}{d x}\right)^3+2 \frac{d y}{d x}=0$
(iii) $\frac{d^2 y}{d x^2}=\sqrt{y-\frac{d y}{d x}}$
(iv) $\frac{d^3 y}{d x^3}=0$
(v) $\frac{d^2 y}{d x^2}+y+\left(\frac{d y}{d x}-\frac{d^3 y}{d x^3}\right)^{\frac{3}{2}}=0$
(vi) $\left(2-y^{\prime \prime}\right)^2=y^{\prime \prime 2}+2 y^{\prime}$
(vii) $\left(\frac{d y}{d x}\right)^3+y=x-\frac{d x}{d y}$
Solution:
(i) $\frac{d y}{d x}+2 y=x^3$
Highest order derivative is $\frac{d y}{d x}$
Power of $\frac{d y}{d x}$ is 1
$
\therefore \text { order }=1 \quad \text { Degree }=1
$
(ii) $\frac{d^3 y}{d x^3}+3\left(\frac{d y}{d x}\right)^3+2 \frac{d y}{d x}=0$
Highest order derivative is $\frac{d^3 y}{d x^3}$
Power of $\frac{d^3 y}{d x^3}$ is 1
$
\therefore \text { order }=3 \quad \text { Degree }=1
$
(iii) $\frac{d^2 y}{d x^2}=\sqrt{y-\frac{d y}{d x}}$

Here we eliminate the radical sign. Squaring both sides we get,
$
\left(\frac{d^2 y}{d x^2}\right)^2=y-\frac{d y}{d x}
$
$
\therefore \text { order }=2, \quad \text { Degree }=2
$

(iv) $\frac{d^3 y}{d x^3}=0$
order $=3, \quad$ Degree $=1$
(v) $\frac{d^2 y}{d x^2}+y+\left(\frac{d y}{d x}-\frac{d^3 y}{d x^3}\right)^{\frac{3}{2}}=0$
Here we eliminate the radical sign
For this write the equation as $\frac{d^2 y}{d x^2}+y=-\left(\frac{d y}{d x}-\frac{d^3 y}{d x^3}\right)^{\frac{3}{2}}$
Squaring both the sides, we get $\left(\frac{d^2 y}{d x^2}+y\right)^2=\left(\frac{d y}{d x}-\frac{d^3 y}{d x^3}\right)^3$
$\therefore$ Order $=3$, Degree $=3$
(vi) $\left(2-y^{\prime \prime}\right)^2=y^{\prime \prime 2}+2 y^{\prime}$
$4-4 y^{\prime \prime}+\left(y^{\prime \prime}\right)^2=\left(y^{\prime \prime}\right)^2+2 y^{\prime}$
$\Rightarrow 4-4 y^{\prime \prime}=2 y^{\prime}$
$\therefore$ order $=2$, Degree $=1$
(vii) $\left(\frac{d y}{d x}\right)^3+y=x-\frac{d x}{d y}$
multiplying the equation by $\frac{d y}{d x}$
$
\begin{aligned}
& \left(\frac{d y}{d x}\right)^4+y \frac{d y}{d x}=x \frac{d y}{d x}-1 \\
& \therefore \text { Order }=1, \text { Degree }=4
\end{aligned}
$

Question 2.
Find the differential equation of the following
(i) $y=c x+c-c^3$
(ii) $y=c(x-c)^2$
(iii) $\mathrm{xy}=\mathrm{c}^2$
(iv) $x^2+y^2=a^2$
Solution:
(i) $y=c x+c-c^3 \ldots \ldots$ (1)
Here $c$ is a constant which has to be eliminated
Differentiating w.r.t $\mathrm{x}, \frac{d y}{d x}=\mathrm{c}$
Using (2) in (1) we get, $y=\left(\frac{d y}{d x}\right) x+\frac{d y}{d x}-\left(\frac{d y}{d x}\right)^3$ which is the required differential equation.
(ii) $y=c(x-c)^2$
We have to eliminate $c$
Differentiating w.r.t $\mathrm{x}$, we get, $\frac{d y}{d x}=2 \mathrm{c}(\mathrm{x}-\mathrm{c})$
Dividing (2) by (1) we get

$
\begin{aligned}
\frac{d y}{d x} & =\frac{2 c(x-c)}{c(x-c)^2} \\
\Rightarrow \quad \frac{d y}{d x} & =\frac{2 y}{x-c} \\
(x-c) \frac{d y}{d x} & =2 y \\
\frac{x d y}{d x}-\frac{c d y}{d x} & =2 y \\
\frac{x d y}{d x}-2 y & =\frac{c d y}{d x} \\
\Rightarrow \quad c & =\frac{\frac{x d y}{d x}-2 y}{\frac{d y}{d x}}
\end{aligned}
$
Using (3) in (1) we get

$
\begin{aligned}
& y=\left[\frac{\frac{x d y}{d x}-2 y}{\frac{d y}{d x}}\right]\left[x-\left(\frac{\frac{x d y}{d x}-2 y}{\frac{d y}{d x}}\right)\right]^2 \\
& y=\frac{\left(\frac{x d y}{d x}-2 y\right)}{\frac{d y}{d x}}\left[\frac{\frac{x d y}{d x}-\frac{x d y}{d x}+2 y}{\left(\frac{d y}{d x}\right)}\right]^2 \\
& \Rightarrow\left(\frac{d y}{d x}\right)^3 y=\left(\frac{x d y}{d x}-2 y\right)\left(4 y^2\right) \\
& \Rightarrow\left(\frac{d y}{d x}\right)^3=4 x y \frac{d y}{d x}+8 y^2 \\
&
\end{aligned}
$
(or) $\left(\frac{d y}{d x}\right)^3-4 x y \frac{d y}{d x}+8 y^2=0$ is the required differential equation
(iii) $x y=c^2$
Differentiating w.r.t $x$, we get, $\frac{x d y}{d x}+y=0$ is the required differential equation
(iv) $x^2+y^2=a^2$
Differentiating w.r.t $x$, we get,
$
2 x+2 y \frac{d y}{d x}=0
$
(or) $x+y \frac{d y}{d x}=0$ is the required differential equation
Question 3 .
Form the differential equation by eliminating $\alpha$ and $\beta$ from $(x-\alpha)^2+(y-\beta)^2=r^2$

Solution:

$
(x-\alpha)^2+(y-\beta)^2=r^2
$
Differentiating w.r.t $x$ we get, $2(x-\alpha)+2(y-\beta) \frac{d y}{d x}=0$ (or) $(x-\alpha)+(y-\beta) \frac{d y}{d x}=0$
Again differentiating w.r.t $x$,
$
\begin{aligned}
1+(y-\beta) \frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^2 & =0 \\
(y-\beta) \frac{d^2 y}{d x^2} & =-1-\left(\frac{d y}{d x}\right)^2
\end{aligned}
$
For convenience we use $y^{\prime}=\frac{d y}{d x}$ and $y^{\prime \prime}=\frac{d^2 y}{d x^2}$
The above equation becomes,
$
\begin{aligned}
(y-\beta) y^{\prime \prime} & =-\left(1+y^{\prime 2}\right) \\
y-\beta & =-\frac{\left(1+y^{\prime 2}\right)}{y^{\prime \prime}}
\end{aligned}
$

Using (3) in (2) we get
$
\begin{array}{r}
(x-\alpha)-\frac{\left(1+y^{\prime 2}\right)}{y^{\prime \prime}} y^{\prime}=0 \\
(\text { or })(x-\alpha)=\frac{\left(1+y^{\prime 2}\right)}{y^{\prime \prime}} y^{\prime}
\end{array}
$
Now using (3) and (4) in (1) we get
$
\begin{aligned}
& {\left[\frac{\left(1+y^{\prime 2}\right)}{y^{\prime \prime}} y^{\prime}\right]^2+\left[\frac{-\left(1+y^{\prime 2}\right)}{y^{\prime \prime}}\right]^2=r^2} \\
& \text { (or) } \begin{aligned}
\frac{\left(1+y^{\prime 2}\right)^2}{\left(y^{\prime \prime}\right)^2}\left(y^{\prime 2}+1\right) & =r^2 \\
\left(1+y^{\prime 2}\right)^3 & =r^2\left(y^{\prime \prime}\right)^2
\end{aligned} \\
& \Rightarrow \quad\left[1+\left(\frac{d y}{d x}\right)^2\right]^3=r^2\left(\frac{d^2 y}{d x^2}\right)^2 \text { is the required differential equation } \\
&
\end{aligned}
$
Question 4.
Find the differential equation of the family of all straight lines passing through the origin. Solution:
The general equation for a family of lines passing through the origin is $\mathrm{y}=\mathrm{mx} \ldots \ldots . .(1)$
Differentiating w.r.t $\mathrm{x}$, $\frac{d y}{d x}=\mathrm{m}$....... (2)
Using (2) in (1)
$\mathrm{y}=\left(\frac{d y}{d x}\right) \mathrm{x}$ is the required differential equation

Question 5.
Form the differential equation that represents all parabolas each of which has a latus rectum $4 \mathrm{a}$ and whose axes are parallel to the $\mathrm{x}$-axis.
Solution:
Equation of parabola whose axis is parallel to the $\mathrm{x}$-axis with latus rectum $4 \mathrm{a}$ is $(y-\beta)^2=4 a(x-\alpha) \ldots \ldots . .(1)$
Here $(\alpha, \beta)$ is the vertex of the parabola.
Differentiating (1) w.r.t $x$, we get
$2(\mathrm{y}-\beta) \frac{d y}{d x}=4 \mathrm{a} \ldots \ldots(2)$
Again, differentiating (2) w.r.t $\mathrm{x}$, we get
$
2\left[(y-\beta) \frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^2\right]=0
$
From (2) we have,
Using this in (3) we get $y-\beta=\frac{2 a}{d y}$
$
\frac{2 a}{\left(\frac{d y}{d x}\right)} \frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^2=0
$
(or) $2 a \frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^3=0$ is the required differential equation

Question 6.
Find the differential equation of all circles passing through the origin and having their centers on the y-axis.
Solution:

The circles pass through the origin. They have their centres at $(0, a)$ The circles have radius a. so the equation of the family of circles is given by $x^2+(y-a)^2=a^2$
$
\begin{aligned}
& x^2+y^2-2 a y+a^2=a^2 \\
& x^2+y^2=2 a y \ldots \ldots .(1)
\end{aligned}
$
Differentiating w.r.t $\mathrm{x}$,

$
\begin{aligned}
& 2 x+2 y \frac{d y}{d x}=2 a \frac{d y}{d x} \\
& \Rightarrow \quad a=\frac{x+y \frac{d y}{d x}}{\frac{d y}{d x}} \ldots \ldots \ldots . \\
& \text { Using (2) in (1) } \\
& x^2+y^2=2 y\left[\frac{x+y \frac{d y}{d x}}{\frac{d y}{d x}}\right] \\
& \Rightarrow \quad\left(x^2-y^2\right) \frac{d y}{d x}=2 x y
\end{aligned}
$
$\frac{d y}{d x}=\frac{2 x y}{x^2-y^2}$ is the required differential equation of all circles passing through origin and having their centres on the $y$-axis.
Question 7.
Find the differential equation of the family of a parabola with foci at the origin and axis along the $\mathrm{x}$-axis.
Solution:
The given family of parabolas have foci at the origin and axis along the $\mathrm{x}$-axis.

The equation of such family of parabolas is given by

$
\mathrm{y}^2=4 \mathrm{a}(\mathrm{x}+\mathrm{a}) \ldots \ldots(1)
$
Differentiating w.r.t $x$,
$
\begin{aligned}
y^2 & =2 y \frac{d y}{d x}\left[x+\frac{y}{2} \frac{d y}{d x}\right] \\
\text { (or) } y & =2 \frac{d y}{d x}\left(x+\frac{y}{2} \frac{d y}{d x}\right)
\end{aligned}
$
$y=2 x \frac{d y}{d x}+y\left(\frac{d y}{d x}\right)^2$ is the required differential equation.

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