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

Updated On May 15, 2024

By SaraNextGen

Additional Problems
I. One Mark Questions
Choose the correct answer.
Question 1.
The differential equation of straight lines passing through the origin is
(a) $\frac{x d y}{d x}=y$
(b) $\frac{d y}{d x}=\frac{x}{y}$
(c) $\frac{d y}{d x}=0$
(d) $\frac{x d y}{d x}=\frac{1}{y}$
Answer:
(a) $\frac{x d y}{d x}=y$
Hint:
$
y=m x ; \frac{d y}{d x}=m \Rightarrow y=\frac{x d y}{d x}
$
Question 2.
The solution of $x d y+y d x=0$ is
(a) $x+y=c$
(b) $x^2+y^2=c$
(c) $x y=c$
(d) $y=c x$
Answer:
(c) $x y=c$
Hint:
$x d y+y d x=0$

$
\begin{aligned}
& d(x y)=0 \\
& x y=c
\end{aligned}
$
Question 3.
The solution of $\mathrm{x} \mathrm{dx}+\mathrm{y} \mathrm{dy}=0$ is
(a) $x^2+y^2=c$
(b) $\frac{x}{y}=\mathrm{c}$
(c) $x^2-y^2=c$
(d) $x y=c$
Answer:
(a) $x^2+y^2=c$
Hint:
$
\begin{aligned}
& \mathrm{xdx}=-\mathrm{ydy} \\
& \frac{x^2}{2}=\frac{-y^2}{2}+c_1 \\
& \mathrm{x}^2+\mathrm{y}^2=\mathrm{c}
\end{aligned}
$
Question 4.
The solution of $\frac{d y}{d x}=\mathrm{e}^{\mathrm{x}-\mathrm{y}}$ is
(a) $e^y e^x=c$
(b) $y=\log \mathrm{ce}^{\mathrm{x}}$
(c) $y=\log \left(e^x+c\right)$
(d) $\mathrm{e}^{x+y}=\mathrm{c}$
Answer:
(c) $y=\log \left(e^x+c\right)$
Hint:
$
\begin{aligned}
\frac{d y}{d x} & =\frac{e^x}{e^y} \\
\int e^y d y & =\int e^x d x \Rightarrow e^y=e^x+c \\
y & =\log \left(e^x+c\right)
\end{aligned}
$
Question 5.
The solution of $\frac{d p}{d t}=\mathrm{ke}^{-\mathrm{t}}$ ( $\mathrm{k}$ is a constant) is
(a) $c-\frac{k}{e^t}=p$
(b) $\mathrm{p}=\mathrm{ke}^{\mathrm{t}}+\mathrm{c}$
(c) $\mathrm{t}=\log \frac{c-p}{k}$

(d) $t=\log _c p$
Answer:
(a) $c-\frac{k}{e^t}=p$
Hint:
$
\begin{aligned}
& d p=k e^{-t} d t \Rightarrow p=-k e^{-t}+c \\
& p=c-\frac{k}{e^t}
\end{aligned}
$
Question 6.
The integrating factor of $\left(1+\mathrm{x}^2\right) \frac{d y}{d x}+\mathrm{xy}=\left(1+\mathrm{x}^2\right)^3$ is
(a) $\sqrt{1+x^2}$
(b) $\log \left(1+x^2\right)$
(c) $e^{\tan ^{-1} x}$
(d) $\log \left(\tan ^{-1} x\right)$
Answer:
(a) $\sqrt{1+x^2}$
Hint:
$
\begin{aligned}
\frac{d y}{d x}+\frac{x}{1+x^2} & =\left(1+x^2\right)^2 \\
\int \frac{x}{1+x^2} & =\frac{1}{2} \log \left(1+x^2\right)=\log \sqrt{1+x^2} \\
\text { I.F } & =e^{\log \sqrt{1+x^2}}=\sqrt{1+x^2}
\end{aligned}
$
Question 7.
The complementary function of the differential equation $\left(\mathrm{D}^2-\mathrm{D}\right) \mathrm{y}=\mathrm{e}^{\mathrm{x}}$ is
(a) $\mathrm{A}+\mathrm{B} \mathrm{e}^{\mathrm{x}}$
(b) $(A x+B) e^x$
(c) $\mathrm{A}+\mathrm{Be}^{-\mathrm{x}}$
(d) $(\mathrm{A}+\mathrm{Bx}) \mathrm{e}^{-\mathrm{x}}$
Answer:
(a) $\mathrm{A}+\mathrm{B} \mathrm{e}^{\mathrm{x}}$
Hint:
$
\begin{aligned}
& \mathrm{m}^2-\mathrm{m}=0 \\
& \mathrm{~m}(\mathrm{~m}-1)=0 \\
& C F=A e^{0 x}+B e^x=A+B e^x
\end{aligned}
$

Question 8.
Match the following

Answer:
(a) - (iii)
(b) - (i)
(c) - (iv)
(d) - (ii)
Question 9.
Fill in the blanks
(a) The general solution of the equation $\frac{d y}{d x}+\frac{y}{x}=1$ is _______________
(b) Integrating factor of $\frac{x d y}{d x}-\mathrm{y}=\sin \mathrm{x}$ is _________________
(c) The differential equation of $y=A \sin x+B \cos x$ is ___________
(d) The D.E $\frac{d y}{d x}+\frac{y}{x \log x}=\frac{1}{x}$ is a ___________ differential equation.
Answer:
(a) $y=\frac{x}{2}+\frac{c}{x}$
(b) $\frac{1}{x}$
(c) $\frac{d^2 y}{d x^2}+y=0$
(d) linear
Question 10.
State true or false
(a) $\mathrm{y}=3 \sin \mathrm{x}+4 \cos \mathrm{x}$ is a particular solution of the differential equation $\frac{d^2 y}{d x^2}+\mathrm{y}=0$
(b) The solution of $\frac{d y}{d x}=\frac{x+2 y}{x}$ is $\mathrm{x}+\mathrm{y}=\mathrm{kx}^2$
(c) $y=13 \mathrm{e}^{\mathrm{x}}+4 \mathrm{e}^{-\mathrm{x}}$ is a solution of $\frac{d^2 y}{d x^2}-\mathrm{y}=0$
Answer:
(a) True
(b) True
(c) True
II. 2 Marks Questions

Question 1.
Form the D.E of the family of curves $y=a e^{3 x}+b e^x$ where $a, b$ are parameters
Solution:
$
\begin{aligned}
& y=a e^{3 x}+b e^x \\
& \frac{d y}{d x}=3 \mathrm{ae}^{3 \mathrm{x}}+\mathrm{be}^{\mathrm{x}} \\
& \frac{d^2 y}{d x^2}=9 a e^{3 x}+b e^x \\
& \frac{d y}{d x}-y=2 a e^{3 x} \\
& \frac{d^2 y}{d x^2}-\frac{d y}{d x}=6 a e^{3 x}=3\left(\frac{d y}{d x}-y\right) \Rightarrow \frac{d^2 y}{d x^2}-4 \frac{d y}{d x}+3 y=0 \\
&
\end{aligned}
$
Question 2.
Find the D.E of a family of curves $y=a \cos (\mathrm{mx}+\mathrm{b})$, $a$ and $\mathrm{b}$ are constants.
Solution:
$
\begin{aligned}
y & =a \cos (m x+b) \\
\frac{d y}{d x} & =-m a \sin (m x+b) \\
\frac{d^2 y}{d x^2} & =-m^2 a \cos (m x+b)=-m^2 y \\
\therefore \frac{d^2 y}{d x^2}+m^2 y & =0 \quad \text { is the required D.E }
\end{aligned}
$
Question 3.
Find the D.E by eliminating the constants a and $b$ from $y=a \tan x+b \sec x$

Solution:
$
\begin{aligned}
y & =a \frac{\sin x}{\cos x}+\frac{b}{\cos x} \\
y \cos x & =a \sin x+b \\
\Rightarrow y(-\sin x)+\cos x \frac{d y}{d x} & =a \cos x \\
\Rightarrow \quad-y \tan x+\frac{d y}{d x} & =a
\end{aligned}
$
Again differentiating w.r.t.x,
$
-y \sec ^2 x-\tan x \frac{d y}{d x}+\frac{d^2 y}{d x^2}=0
$
Question 4.
Solve: $\frac{d y}{d x}=e^{7 x+y}$
Solution:
$
\begin{aligned}
\frac{d y}{d x} & =e^{7 x} e^y \Rightarrow \frac{d y}{e^y}=e^{7 x} d x \\
\Rightarrow \int e^{-y} d y & =\int e^{7 x} d x+c \\
\Rightarrow-e^{-y} & =\frac{e^{7 x}}{7}+c \Rightarrow \frac{e^{7 x}}{7}+e^{-y}=c
\end{aligned}
$
Question 5.
Solve: $\left(x^2-a y\right) d x=\left(a x-y^2\right) d y$
Solution:
Writing the equation as
$
\begin{aligned}
& x^2 d x+y^2 d y=a(x d y+y d x) \\
& x^2 d x+y^2 d y=a d(x y) \\
& \int x^2 d x+\int y^2 d y=a \int d(x y)+c \\
& \frac{x^3}{3}+\frac{y^3}{3}=a x y+c
\end{aligned}
$
Hence the general solution is $x^3+y^3=3 a x y+c$

Question 6.
Solve $(\sin x+\cos x) d y+(\cos x-\sin x) d x=0$
Solution:
The given equation can be written as
$
\begin{aligned}
& d y+\frac{\cos x-\sin x}{\sin x+\cos x} d x=0 \\
\Rightarrow & \int d y+\int \frac{\cos x-\sin x}{\sin x+\cos x} d x=c \\
\Rightarrow & y+\log (\sin x+\cos x)=c
\end{aligned}
$
III. 3 and 5 Marks Questions
Question 1.
Solve $\frac{x d y}{d x}+\cos \mathrm{y}=0$, given $\mathrm{y}=\frac{\pi}{4}$ when $\mathrm{x}=\sqrt{ } 2$
Solution:
$
\begin{aligned}
x d y & =-\cos y d x \\
\therefore \int \sec y d y & =-\int \frac{d x}{x}+\log c \\
\log (\sec y+\tan y)+\log x & =\log c \\
\text { or } \quad x(\sec y+\tan y) & =c \\
\text { Put } x=\sqrt{2} \text { and } y & =\frac{\pi}{4} \\
\sqrt{2}\left(\sec \frac{\pi}{4}+\tan \frac{\pi}{4}\right) & =c \\
\text { (or } c=\sqrt{2}(\sqrt{2}+1) & =2+\sqrt{2}
\end{aligned}
$
$\therefore$ The particular solution is $x(\sec y+\tan y)=2+\sqrt{2}$

Question 2.
The slope of a curve at any point is the reciprocal of twice the ordinate of the point. The curve also passes through the point $(4,3)$. Find the equation of the curve.
Solution:
Slope at any point $(x, y)$ is the slope of the tangent at $(x, y)$
$
\begin{aligned}
& \frac{d y}{d x}=\frac{1}{2 y} \\
& \Rightarrow 2 \mathrm{y} \mathrm{dy}=\mathrm{dx} \\
& \Rightarrow \int 2 \mathrm{y} \mathrm{dy}=\int \mathrm{dx}+\mathrm{c} \\
& \Rightarrow \mathrm{y}^2=\mathrm{x}+\mathrm{c}
\end{aligned}
$
Since the curve passes through $(4,3)$
we have $9=4+\mathrm{c} \Rightarrow \mathrm{c}=5$
Equation of the curve is $y^2=x+5$
Question 3.
The net profit $\mathrm{P}$ and quantity $\mathrm{x}$ satisfy the differential equation $\frac{d p}{d x}=\frac{2 p^3-x^3}{3 x p^2}$. Find the relationship between the net profit and demand given that $p=20$ when $x=10$
Solution:
$\frac{d p}{d x}=\frac{2 p^3-x^3}{3 x p^2}$ is a homogeneous differential equation
Put $p=v x$ and $\frac{d p}{d x}=v+x \frac{d v}{d x}$
$
\begin{aligned}
v+x \frac{d v}{d x} & =\frac{2 v^3-1}{3 v^2} \\
\Rightarrow \quad x \frac{d v}{d x} & =\frac{2 v^3-1}{3 v^2}-v=\frac{2 v^3-3 v^3-1}{3 v^2}
\end{aligned}
$

$
\begin{aligned}
x \frac{d v}{d x} & =\frac{-\left(1+v^3\right)}{3 v^2} \\
\int \frac{3 v^2}{1+v^3} d v & =-\int \frac{d x}{x} \\
\log \left(1+v^3\right) & =\log \frac{1}{x}+\log c \\
\log \left(1+v^3\right) & =\log \frac{c}{x} \Rightarrow(\text { i.e }) 1+v^3=\frac{c}{x} \\
\text { Replace } \quad v & =\frac{p}{x} ; x^3+p^3=c x^2
\end{aligned}
$
When $x=10, p=20$
$
\begin{aligned}
\therefore(10)^3+(20)^3 & =c(10)^2 \\
1000+8000 & =c(100) \Rightarrow c=90 \\
\therefore x^3+p^3 & =90 x^2 \\
p^3 & =x^2(90-x) \text { is the required relationship }
\end{aligned}
$

Question 4.
Solve: $\frac{d y}{d x}+\mathrm{ay}=\mathrm{e}^{\mathrm{x}}(\mathrm{a} \neq-1)$
Solution:
The given equation is of the form $\frac{d y}{d x}+P y=Q$
Here $\mathrm{P}=\mathrm{a}, \mathrm{Q}=\mathrm{e}^{\mathrm{x}}$
The general solution is $y$ (I.F) $=\int \mathrm{Q}$ (I.F) $d x+c$
$
\begin{aligned}
\text { I.F } & =e^{\int p d x}=e^{\int a d x}=e^{a x} \\
\Rightarrow \quad y e^{a x} & =\int e^x e^{a x} d x+c \\
& =\int e^{(a+1) x} d x+c \\
\Rightarrow \quad y e^{a x} & =\frac{e^{(a+1) x}}{a+1}+c
\end{aligned}
$
Question 5.
Solve: $\frac{d y}{d x}+\mathrm{y} \cos \mathrm{x}=\frac{1}{2} \sin 2 \mathrm{x}$
Solution:
Here $P=\cos \mathrm{x}, \mathrm{Q}=\frac{1}{2} \sin 2 \mathrm{x}$
$
\begin{aligned}
\int p d x & =\int \cos x d x=\sin x \\
\text { I.F } & =e^{\int p d x}=e^{\sin x}
\end{aligned}
$
The general solution is
$
\begin{aligned}
y e^{\sin x} & =\int \frac{1}{2} \sin 2 x e^{\sin x} d x+c \\
& =\int \sin x \cos x e^{\sin x} d x+c \\
& =\int t e^t d t+c(\mathrm{t}=\sin x) \\
& =e^t(t-1)+c \text { (Using integration by parts) } \\
y e^{\sin x} & =e^{\sin x}(\sin x-1)+c
\end{aligned}
$
Question 6.
Solve $\left(D^2-6 D+25\right) y=0$
Solution:
The auxiliary equations is $\mathrm{m}^2-6 \mathrm{~m}+25=0$

$
m=\frac{6 \pm \sqrt{36-100}}{2}=\frac{6 \pm 8 i}{2}=3 \pm 4 i
$
The Roots are complex and of the form, $\alpha \pm \beta$ with $\alpha=3$ and $\beta=4$
The complementary function $=e^{3 x}(A \cos 4 x+B \sin 4 x)$
The general solution is $y=e^{3 x}(A \cos 4 x+B \sin 4 x)$
Question 7.
Solve $\left(D^2+10 D+25\right) y=\frac{5}{2}+e^{-5 x}$
Solution:
The auxiliary equations is $\mathrm{m}^2+10 \mathrm{~m}+25=0$
$
\begin{aligned}
& (\mathrm{m}+5)^2=0 \\
& \mathrm{~m}=-5,-5
\end{aligned}
$
The complementary function $=(\mathrm{Ax}+\mathrm{B}) \mathrm{e}^{-5 \mathrm{x}}$
$
\begin{aligned}
\text { P.I }_1 & =\frac{1}{\mathrm{D}^2+10 \mathrm{D}+25}\left(\frac{5}{2} e^{o x}\right)=\frac{1}{25}\left(\frac{5}{2}\right)=\frac{1}{10} \\
\mathrm{P.I}_2 & =\frac{1}{\mathrm{D}^2+10 \mathrm{D}+25} e^{-5 x}=\frac{1}{(\mathrm{D}+5)^2} e^{-5 x} \\
& =x \frac{1}{2(\mathrm{D}+5)} e^{-5 x}=\frac{x^2}{2} e^{-5 x}
\end{aligned}
$
The general solution is
$
\begin{aligned}
& y=\mathrm{C} \cdot \mathrm{F}+\mathrm{P}_1+\mathrm{I}_1+\mathrm{P}_2 \\
& y=(\mathrm{A} x+\mathrm{B}) e^{-5 x}+\frac{1}{10}+\frac{x^2}{2} e^{-5 x}
\end{aligned}
$
Question 8.
Suppose that the quantity demanded $\mathrm{Q}_{\mathrm{d}}=40-4 \mathrm{p}-4 \frac{d p}{d t}+\frac{d^2 p}{d t^2}$ and quantity supplied $\mathrm{Q}_{\mathrm{S}}=-6+$ $8 p$ where $p$ is the price. Find the equilibrium price for market clearance.
Solution:

For market clearance, the required condition is $\mathrm{Q}_{\mathrm{d}}=\mathrm{Q}_{\mathrm{s}}$
$
\Rightarrow \quad \begin{aligned}
40-4 p-4 \frac{d p}{d t}+\frac{d^2 p}{d t^2} & =-6+8 p \\
\text { (or) } \frac{d^2 p}{d t^2}-4 \frac{d p}{d t}-12 p & =-46
\end{aligned}
$
The auxiliary equation is $m^2-4 m-12=0$
$
\begin{aligned}
(m-6)(m+2) & =0 \Rightarrow m=6,-2 \\
\text { C.F } & =\mathrm{A} e^{6 t}+\mathrm{B} e^{-2 t} \\
\text { P.I. } & =\frac{1}{\mathrm{D}^2-4 \mathrm{D}-12}(-46) e^{0 t}=\frac{-1}{12}(-46)=\frac{23}{6}
\end{aligned}
$
The general solution is $P=$ C.F $+$ P.I
$
P=\mathrm{A} e^{6 t}+\mathrm{B} e^{-2 t}+\frac{23}{6}
$