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

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Miscellaneous Problems
Question 1.
Suppose that $\mathrm{Q}_{\mathrm{d}}=30-5 \mathrm{P}+2 \frac{d p}{d t}+\frac{d^2 p}{d t^2}$ and $\mathrm{Q}_{\mathrm{S}}=6+3 \mathrm{P}$. Find the equilibrium price for market clearance.
Solution:
For equilibrium price $\mathrm{Q}_{\mathrm{d}}=\mathrm{Q}_{\mathrm{S}}$
$
\begin{gathered}
\Rightarrow 30-5 p+2 \frac{d p}{d t}+\frac{d^2 p}{d t^2}=6+3 p \\
\Rightarrow \quad \frac{d^2 p}{d t^2}+2 \frac{d p}{d t}-8 p=-24 \\
\text { (or) } \quad\left(\mathrm{D}^2+2 \mathrm{D}-8\right) p=-24, \quad \mathrm{D}=\frac{d}{d t}
\end{gathered}
$
The auxiliary equation is $m^2+2 m-8=0$
$\therefore \quad$ C.F is $\mathrm{A} e^{-4 t}+\mathrm{B} e^{2 t}$
$
P . I=\frac{1}{\mathrm{D}^2+2 \mathrm{D}-8}\left(-24 e^{o t}\right)=\frac{-24}{0+0-8}=3
$
Hence the equilibrium price
$
P=C F+P I
$
$
\mathrm{P}=\mathrm{A} e^{-4 t}+\mathrm{B} e^{2 t}+3
$
Question 2.
Form the differential equation having for its general solution $y=a x^2+b x$ Solution:
Given $y=a x^2+b x$
Since there are 2 constants a, b we have to differentiate twice to eliminate them
$
\begin{aligned}
y & =a x^2+b x \\
\frac{d y}{d x} & =2 a x+b \\
\frac{d^2 y}{d x^2} & =2 a \\
\Rightarrow a & =\frac{1}{2} \frac{d^2 y}{d x^2}
\end{aligned}
$
Using this in (2) we get

$
\frac{d y}{d x}=x \frac{d^2 y}{d x^2}+b \quad \Rightarrow b=\frac{d y}{d x}-x \frac{d^2 y}{d x^2}
$
Use this in (1)
$
\begin{aligned}
& y=\frac{1}{2} \frac{d^2 y}{d x^2} x^2+x\left[\frac{d y}{d x}-x \frac{d^2 y}{d x^2}\right] \\
& 2 y=x^2 \frac{d^2 y}{d x^2}+2 x \frac{d y}{d x}-2 x^2 \frac{d^2 y}{d x^2} \\
& 2 y=-2 x^2 \frac{d^2 y}{d x^2}+2 x \frac{d y}{d x} \\
& \text { (or) } x^2 \frac{d^2 y}{d x^2}-2 x \frac{d y}{d x}+2 y=0 \text { is the required differential solution }
\end{aligned}
$
Question 3.
Solve $y x^2 d x+e^{-x} d y=0$
Solution:
The given equation can be written as
$
\begin{aligned}
e^{-x} d y & =-y x^2 d x \\
\frac{d y}{y} & =\frac{-x^2}{e^{-x}} d x \Rightarrow \frac{d y}{y}=-x^2 e^x d x
\end{aligned}
$
Integrating,
$
\begin{aligned}
& \int \frac{d y}{y}=-\int x^2 e^x d x \\
& \log y=-\left[x^2 e^x-\int 2 x e^x d x\right]=-x^2 e^x+2 \int x e^x d x \\
& =-x^2 e^x+2\left[x e^x-\int e^x d x\right] \\
& \log y=-x^2 e^x+2 x e^x-2 e^x+c \\
& \text { (or) } e^x\left(x^2-2 x+2\right)-\log y=c \text { is the required solution } \\
&
\end{aligned}
$

Question 4.
Solve: $\left(x^2+y^2\right) d x+2 x y d y=0$
Solution:
The given equation can be written as $\frac{d y}{d x}=-\frac{\left(x^2+y^2\right)}{2 x y}$ It is a homogeneous differential equation
$
\begin{aligned}
& \text { Put } y=v x \text { and } \quad \frac{d y}{d x}=v+x \frac{d v}{d x} \\
& v+x \frac{d v}{d x}=-\frac{\left(x^2+v^2 x^2\right)}{2 x(x v)}=\frac{-\left(1+v^2\right)}{2 v} \\
& x \frac{d v}{d x}=\frac{-1-v^2}{2 v}-v \\
& =\frac{-1-v^2-2 v^2}{2 v}=\frac{-1-3 v^2}{2 v} \\
&
\end{aligned}
$
Separating the variables,
$
\begin{aligned}
\frac{2 v}{1+3 v^2} & =-\frac{d x}{x} \\
\text { Integrating, } \int \frac{2 v}{1+3 v^2} & =-\int \frac{d x}{x} \\
\frac{1}{3} \log \left(1+3 v^2\right) & =-\log x+\log c \\
\Rightarrow \log \left(1+3 v^2\right)^{\frac{1}{3}} & =\log \left(\frac{1}{x}\right)+\log c
\end{aligned}
$

$
\begin{aligned}
& \Rightarrow \log \left(1+3 v^2\right)^{\frac{1}{3}}=\log \left(\frac{1}{x}\right)+\log c \\
& \left(1+3 v^2\right)^{\frac{1}{3}}=\frac{c}{x} \\
& \text { Replace } \quad v=\frac{y}{x} \\
& \left(1+3 \frac{y^2}{x^2}\right)^{\frac{1}{3}}=\frac{c}{x} \quad \text { (or) } x\left(1+3 \frac{y^2}{x^2}\right)^{\frac{1}{3}}=c \text { is the required solution } \\
&
\end{aligned}
$
Question 5.
Solve: $\mathrm{x} \frac{d y}{d x}+2 \mathrm{y}=\mathrm{x}^4$
Solution:
The given equation can be written as $\frac{d y}{d x}+\frac{2}{x} y=x^3$
This is of the form $\frac{d y}{d x}+\mathrm{Py}=\mathrm{Q}$
Where $\mathrm{P}=\frac{2}{x}$ and $\mathrm{Q}=x^3$
$
\text { Now } \begin{aligned}
\int p d x & =\int \frac{2}{x} d x=2 \log x=\log x^2 \\
\text { I.F } & =e^{\int p d x}=e^{\log x^2}=x^2
\end{aligned}
$
The solution is $y$ (I.F) $=\int \mathrm{Q}($ I.F $) d x+c$
$
y x^2=\int x^5 d x+c \quad \text { (or) } y x^2=\frac{x^6}{6}+c
$

Question 6.
A manufacturing company has found that the cost $\mathrm{C}$ of operating and maintaining the equipment is related to the length ' $m$ ' of intervals between overhauls by the equation $m^2 \frac{d C}{d m}$ $+2 \mathrm{mC}=2$ and $\mathrm{c}=4$ and when $\mathrm{m}=2$. Find the relationship between $\mathrm{C}$ and $\mathrm{m}$.
Solution:
$
\begin{array}{r}
m^2 \frac{d C}{d m}+2 m \mathrm{C}=2 \\
\text { or } \quad \frac{d C}{d m}+\frac{2}{m} \mathrm{C}=\frac{2}{m^2}
\end{array}
$
This is of the form $\frac{d C}{d m}+\mathrm{P} y=\mathrm{Q}$
Where $\mathrm{P}=\frac{2}{m}$ and $\mathrm{Q}=\frac{2}{m^2}$
Now $\quad \int P d m=\int \frac{2}{m} d m=2 \log m=\log m^2$
$
\text { I.F }=e^{\int p d m}=e^{\log m^2}=m^2
$
The solution is $\mathrm{C}$ (I.F) $=\int \mathrm{Q}$ ( I.F) $d m+k$
$
\begin{aligned}
\mathrm{Cm}^2 & =\int\left(\frac{2}{m^2}\right) m^2 d m+k \\
\mathrm{Cm} & =2 m+k
\end{aligned}
$
given that $\mathrm{c}=4$ when $\mathrm{m}=2$
$
\begin{aligned}
& 4(4)=2(2)+\mathrm{k} \\
& \mathrm{k}=12
\end{aligned}
$
So the relation ship between $\mathrm{C}$ and $\mathrm{m}$ is $\mathrm{Cm}^2=2 \mathrm{~m}+12=2(\mathrm{~m}+6)$
Question 7.
Solve $\left(D^2-3 D+2\right) y=e^{4 x}$ given $y=0$ when $x=0$ and $x=1$.
Solution:
$
\left(D^2-3 D+2\right) y=e^{4 x}
$
The auxiliary equations is $\mathrm{m}^2-3 m+2=0$
$
\begin{aligned}
& (\mathrm{m}-2)(\mathrm{m}-1)=0 \\
& \mathrm{~m}=2,1
\end{aligned}
$

C.F is $\mathrm{A} e^{2 x}+\mathrm{B} e^x$
$
P . I=\frac{1}{\mathrm{D}^2-3 \mathrm{D}+2} e^{4 x}=\frac{e^{4 x}}{16-12+2}=\frac{e^{4 x}}{6}
$
So the general solution is $y=$ C.F + P.I
$
y=\mathrm{Ae}^{2 x}+\mathrm{B} e^x+\frac{e^{4 x}}{6}
$
We have to find the constants A and $\mathrm{B}$ with the given conditions
Given $y=0$, when $x=0$
Given $y=0$, when $x=1$
$
0=A+B+\frac{1}{6}
$
So
(1) $\mathrm{X} e^2$ gives
$
0=\mathrm{A} e^2+\mathrm{B} e+\frac{e^4}{6}
$
Subtracting,
$
\begin{aligned}
& 0=\mathrm{A} e^2+\mathrm{B} e^2+\frac{e^2}{6} \\
& 0=\mathrm{A} e^2+\mathrm{B} e+\frac{e^4}{6}
\end{aligned}
$
$0=\mathrm{B}\left(e^2-e\right)+\frac{e^2}{6}-\frac{e^4}{6}$
$
\mathrm{B}\left(e^2-e\right)=\frac{e^4}{6}-\frac{e^2}{6}
$
$
\mathrm{B} e(e-1)=\frac{e^2}{6}\left(e^2-1\right)
$
$
\mathrm{B} e(e-1)=\frac{e^2}{6}(e-1)(e+1)
$
$
\Rightarrow \quad \mathrm{B}=\frac{e}{6}(e+1)=\frac{e^2}{6}+\frac{e}{6}
$
From (1)
$
\mathrm{A}=-\mathrm{B}-\frac{1}{6}
$

$
\mathrm{A}=\frac{-e^2}{6}-\frac{e}{6}-\frac{1}{6}
$
Using the values of A and B in the general solution, we get
$
y=\left(\frac{-e^2}{6}-\frac{e}{6}-\frac{-1}{6}\right) e^{2 x}+\left(\frac{e^2}{6}+\frac{e}{6}\right) e^x+\frac{e^{4 x}}{6}
$
(or)
$
6 y=\left(e^2+e\right) e^x-\left(e^2+e+1\right) e^{2 x}+e^{4 x}
$
Question 8.
Solve: $\frac{d y}{d x}+\mathrm{y} \cos \mathrm{x}=2 \cos \mathrm{x}$
Solution:
The given equation can be written as $\frac{d y}{d x}+(\cos \mathrm{x}) \mathrm{y}=2 \cos \mathbf{x}$
It is of the form $\frac{d y}{d x}+\mathrm{Py}=\mathrm{Q}$
Where $P=\cos x, Q=2 \cos x$
Now $\int P d x=\int \cos x d x=\sin x$
$
\begin{aligned}
\text { I.F } & =e^{\int p d x}=e^{\sin x} \\
y e^{\sin x} & =\int e^{\sin x}(2 \cos x) d x+c
\end{aligned}
$
(or) $y e^{\sin x}=2 \int e^{\sin x} d(\sin x)+c$
$\Rightarrow \quad y e^{\sin x}=2 e^{\sin x}+c$
Question 9.
Solve: $x^2 y d x-\left(x^3+y^3\right) d y=0$
Solution:
$
\frac{d y}{d x}=\frac{x^2 y}{x^3+y^3}
$
This is homogeneous differential equation
Put $y=v x$ and $\frac{d y}{d x}=\frac{x d v}{d x}+v$
$
\begin{aligned}
\Rightarrow \quad v+x \frac{d v}{d x} & =\frac{x^2(x v)}{x^3+x^3 v^3}=\frac{v}{1+v^3} \\
x \frac{d v}{d x} & =\frac{v}{1+v^3}-v=\frac{v-v-v^4}{1+v^3} \\
x \frac{d v}{d x} & =\frac{-v^4}{1+v^3}
\end{aligned}
$

$\Rightarrow \quad \frac{1+v^3}{v^4}=-\frac{d x}{x}$ Integrating,
$
\int \frac{1+v^3}{v^4} d v=-\int \frac{d x}{x}
$
(or)
$
\begin{aligned}
\int \frac{1}{v^4} d v+\int \frac{1}{v} d v & =-\int \frac{d x}{x} \\
\frac{-1}{3 v^3}+\log v & =-\log x+c
\end{aligned}
$
Replace $\quad v=\frac{y}{x}$
$
\begin{aligned}
& \frac{-x^3}{3 y^3}+\log \frac{y}{x}=\log \frac{1}{x}+c \\
& \frac{-x^3}{3 y^3}+\log y=c
\end{aligned}
$
$\log y=\frac{x^3}{3 y^3}+c$ is the required solution
Question 10.
Solve: $\frac{d y}{d x}=\mathrm{xy}+\mathrm{x}+\mathrm{y}+1$
Solution:
$
\begin{aligned}
\frac{d y}{d x} & =(1+x)(1+y) \\
\frac{d y}{(1+y)} & =(1+x) d x
\end{aligned}
$
Integrating, $\quad \int \frac{d y}{(1+y)}=\int(1+x) d x$
(or)
$\log |1+y|=x+\frac{x^2}{x}+c$ is the required solution