Exercise 10.4 - 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 $10.4$
Question 1.
Show that each of the following expressions is a solution of the corresponding given differential equation.
(i) $y=2 x^2 ; x y \prime=2 y$
Solution:
$\mathrm{v}=2 \mathrm{x}^2$
Differential equation: $x y$ ' $=2 \mathrm{y}$
Differentiate with respect to ' $x$ '
$
y^{\prime}=4 x \Rightarrow x=\frac{y^{\prime}}{4}
$
Substitute in (1)
$
\begin{aligned}
& y=2 x(x) \\
& y=2 x\left(\frac{y^{\prime}}{4}\right) \Rightarrow=\frac{x y^{\prime}}{2}
\end{aligned}
$
On simplifying, $2 \mathrm{y}=\mathrm{xy}$ '
$\therefore(1)$ is solution of the given differential equation.
(ii) $y=a e^x+b e^{-x} ; y^{\prime \prime}-y=0$
Solution:
$\mathrm{y}=\mathrm{ae}^{\mathrm{x}}+\mathrm{be} \mathrm{e}^{-\mathrm{x}} \ldots$ (1) Differential equation: $\mathrm{y}^{\prime \prime}-\mathrm{y}=0$
Differentiate with respect to ' $x$ '
$
\mathrm{y}^{\prime}=a e^{\mathrm{x}}-b \mathrm{e}^{-\mathrm{x}}
$
Again differentiate with respect to ' $x$ '
$
\begin{aligned}
& y^{\prime \prime}=a e^x+b e^{-x} \\
& y "=y \Rightarrow y "-y=0
\end{aligned}
$
$\therefore(1)$ is the solution of the given differential equation.
Question 2.
Find value of $\mathrm{m}$ so that the function $\mathrm{y}=\mathrm{e}^{\mathrm{mx}}$ is a solution of the given differential equation.
(i) $y+2 y=0$
Solution:
Given solution $\mathrm{y}=\mathrm{e}^{\mathrm{mx}}$
Differentiate with respect to ' $x$ '
$
y^{\prime}=m e^{m x} \Rightarrow y^{\prime}=m y
$

$
y^{\prime}-m y=0
$
Comparing with $y^{\prime}+2 y=0 \Rightarrow m=-2$
$
\begin{aligned}
& \text { (ii) } y^{\prime \prime}-5 y^{\prime}+6 \mathrm{y}=0 \\
& y=e^{m x} \\
& y^{\prime}=m e^{m x} \\
& y^{\prime \prime}=m^2 e^{m x}
\end{aligned}
$
Given differential equation is $y^{\prime \prime}-5 y$ ' $+6 y=0$
Substitute (1), (2) and (3) in this
$
\begin{aligned}
m^2 e^{m x}-5 m e^{m x}+6 e^{m x} & =0 \\
\left(\div e^{m x}\right) \Rightarrow m^2-5 m+6 & =0 \\
(m-3)(m-2) & =0 \Rightarrow m=3,2
\end{aligned}
$
Question 3.
The slope of the tangent to the curve at any point is the reciprocal of four times the ordinate at that point. The curve passes through $(2,5)$. Find the equation of the curve.
Solution:
Slope of the tangent is the reciprocal of four times the ordinate
$
\text { i.e., } \begin{aligned}
\frac{d y}{d x} & =\frac{1}{4 y} \\
4 \int y d y & =\int d x \\
4 \frac{y^2}{2} & =x+c \Rightarrow 2 y^2=x+c
\end{aligned}
$
Passes through $(2,5)$
$
\therefore c=50-2=48
$

Question 4.
Show that $y=e^{-x}+m x+n$ is a solution of the differential equation $e^x\left(\frac{d^2 y}{d x^2}\right)-1=0$.
Solution:
Given solution is $y=e^{-x}+m x+x$
Differentiating (1) with respect to ' $x$ '
$
\frac{d y}{d x}=-e^{-x}+m
$
Again, differentiating with respect to ' $x$ '
$
\begin{aligned}
\frac{d^2 y}{d x^2} & =e^{-x}=\frac{1}{e^x} \\
e^x\left(\frac{d^2 y}{d x}\right) & =1 \Rightarrow e^x\left(\frac{d^2 y}{d x^2}\right)-1=0
\end{aligned}
$
Hence proved.

Question 5.
Show that $\mathrm{y}=\mathrm{ax}+\frac{b}{x}, \mathrm{x} \neq 0$, is a solution of the differential equation $\mathrm{x}^2 \mathrm{y}^{\prime \prime}+\mathrm{xy} \prime-\mathrm{y}=0$.
Solution:
Given solution : $y=a x+\frac{b}{x}, x \neq 0$
Here ' $a$ ' and ' $b$ ' are arbitrary constants
$
\begin{array}{r}
y=a x+\frac{b}{x} \\
x y=a x^2+b
\end{array}
$
Differentiate with respect to ' $x$ ' xy' $+\mathrm{y} \cdot 1=\mathrm{a}(2 \mathrm{x})=2 \mathrm{ax} \ldots \ldots \ldots(2)$
Differentiate again with respect to ' $x$ '.
$
x y^{\prime \prime}+y^{\prime} .1+y=2 a \Rightarrow x y^{\prime \prime}+2 y '=2 a
$
Substitute (3) in (2)
$
\begin{aligned}
& x y^{\prime}+y=\left(x y^{\prime \prime}+2 y^{\prime}\right) x \\
& x y^{\prime}+y=x^2 y^{\prime \prime}+2 x y^{\prime} \Rightarrow x^2 y^{\prime \prime}+x y^{\prime}-y=0
\end{aligned}
$
Hence proved.

Question 6.
Show that $\mathrm{y}=\mathrm{ae} \mathrm{e}^{-3 \mathrm{x}}+\mathrm{b}$, where $a$ and $\mathrm{b}$ are arbitrary constants, is a solution of the differential equation
$
\frac{d^2 y}{d x^2}+3 \frac{d y}{d x}=0 \text {. }
$
Solution:
$
y=a e^{-3 x}+b\left({ }^{\prime} a^{\prime} \text { and } b^{\prime} b^{\prime}\right. \text { are arbitrary constants) }
$
Differentiate with respect to ' $x$ '
$
\frac{d y}{d x}=-3 a e^{-3 x}
$
Differentiate again with respect to ' $x$ '
$
\begin{aligned}
\frac{d^2 y}{d x^2} & =-3\left(-3 a e^{-3 x}\right) \\
\frac{d^2 y}{d x^2} & =-3\left(\frac{d y}{d x}\right) \\
\frac{d^2 y}{d x^2}+3 \frac{d y}{d x} & =0
\end{aligned}
$

Question 7.
$
y^2=2 a\left(x+a^{\frac{2}{3}}\right)
$
where $a$ is a
Show that the differential equation representing the family of curves
positive parameter, is

Solution:
$
\left(y^2-2 x y \frac{d y}{d x}\right)^3=8\left(y \frac{d y}{d x}\right)^5
$
Given equation is $y^2=2 a\left(x+a^{2 / 3}\right),{ }^{\prime} a$ ' is the parameter.
$
y^2=2 a x+2 a^{5 / 3}
$
Differentiate with respect to ' $x$ '
$
\begin{aligned}
2 y \frac{d y}{d x} & =2 a \Rightarrow y \frac{d y}{d x}=a \\
(1) \Rightarrow \quad y^2-2 a x & =2 a^{5 / 3}
\end{aligned}
$
Substitute $a=y \frac{d y}{d x}$
$
y^2-2 x y \frac{d y}{d x}=2\left(y \frac{d y}{d x}\right)^{5 / 3}
$
Raising the power both sides to ' 3 '
$
\left(y^2-2 x y \frac{d y}{d x}\right)^3=8\left(y \frac{d y}{d x}\right)^5
$
Hence proved.

Question 8.
Show that, $\mathrm{y}=\mathrm{a} \cos \mathrm{bx}$ is a solution of the differential equation $\frac{d y}{d x}=-a b \sin b x$

Solution:

$\mathrm{y}=\mathrm{a} \cos \mathrm{bx}$...(1) (a is an arbitrary constant) Differentiating with respect to ' $\mathrm{x}$ '
$
\frac{d y}{d x}=-a b \sin b x
$
Again, differentiating with respect to ' $x$ '
$
\begin{aligned}
\frac{d^2 y}{d x^2} & =-a b^2 \cos b x=-b^2(a \cos b x)=-b^2 y \\
\Rightarrow \frac{d^2 y}{d x^2}+b^2 y & =0
\end{aligned}
$
Hence proved