Additional Problems - Chapter 4 - Differential Equations - 12th Maths Guide Guide Samacheer Kalvi Solutions

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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}
$