Exercise 4.5 - Chapter 4 Differential Equations 12th Maths Guide Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Ex $4.5$
Solve the following differential equations:
Question 1.
$
\frac{d^2 y}{d x^2}-6 \frac{d y}{d x}+8 y=0
$
Solution:
Given $\left(\mathrm{D}^2-6 \mathrm{D}+8\right) \mathrm{y}=0, \mathrm{D}=\frac{d}{d x}$
The auxiliary equations is
$
\begin{aligned}
& m^2-6 m+8=0 \\
& (m-4)(m-2)=0 \\
& m=4,2
\end{aligned}
$
Roots are real and different
The complementary function (C.F) is $\left(\mathrm{Ae}^{4 \mathrm{x}}+\mathrm{Be}^{2 \mathrm{x}}\right)$
The general solution is $\mathrm{y}=\mathrm{Ae}^{4 \mathrm{x}}+\mathrm{Be}^{2 \mathrm{x}}$
Question 2.
$
\frac{d^2 y}{d x^2}-4 \frac{d y}{d x}+4 y=0
$
Solution:
The auxiliary equations A.E is $\mathrm{m}^2-4 \mathrm{~m}+4=0$
$
\begin{aligned}
& (\mathrm{m}-2)^2=0 \\
& \mathrm{~m}=2,2
\end{aligned}
$
Roots are real and equal
The complementary function (C.F) is $(\mathrm{Ax}+\mathrm{B}) \mathrm{e}^{2 \mathrm{x}}$
The general solution is $y=(A x+B) e^{2 x}$
Question 3.
$
\left(\mathrm{D}^2+2 \mathrm{D}+3\right) \mathrm{y}=0
$
Solution:
The auxiliary equations A.E is $\mathrm{m}^2+2 \mathrm{~m}+3=0$
$
\begin{aligned}
& \Rightarrow \mathrm{m}^2+2 \mathrm{~m}+1+2=0 \\
& \Rightarrow(\mathrm{m}+1)^2=-2 \\
& \Rightarrow \mathrm{m}+1=\pm \sqrt{ } 2 \mathrm{i} \\
& \Rightarrow \mathrm{m}=-1 \pm \sqrt{ } 2 \mathrm{i}
\end{aligned}
$
It is of the form $\alpha \pm \mathrm{i} \beta$

The complementary function (C.F) $=e^{-x}[A \cos \sqrt{2} x+B \sin \sqrt{2} x]$
The general solution is $y=e^{-x}[A \cos \sqrt{ } 2 x+B \sin \sqrt{2} x]$
Question 4.
$
\frac{d^2 y}{d x^2}-2 k \frac{d y}{d x}+k^2 y=0
$
Solution:
Given $\left(\mathrm{D}^2-2 \mathrm{kD}+\mathrm{k}^2\right) \mathrm{y}=0, \mathrm{D}=\frac{d}{d x}$
The auxiliary equations is $\mathrm{m}^2-2 \mathrm{~km}+\mathrm{k}=0$
$
\begin{aligned}
& \Rightarrow(\mathrm{m}-\mathrm{k})^2=0 \\
& \Rightarrow \mathrm{m}=\mathrm{k}, \mathrm{k}
\end{aligned}
$
Roots are real and equal
The complementary function (C.F) is $(\mathrm{Ax}+\mathrm{B}) \mathrm{e}^{\mathrm{kx}}$
The general solution is $\mathrm{y}=(\mathrm{Ax}+\mathrm{B}) \mathrm{e}^{\mathrm{kx}}$
Question 5.
$\left(\mathrm{D}^2-2 \mathrm{D}-15\right) \mathrm{y}=0$ given that $\frac{d y}{d x}=0$ and $\frac{d^2 y}{d x^2}=2$ when $\mathrm{x}=0$
Solution:
A.E is $m^2-2 m-15=0$
$
(\mathrm{m}-5)(\mathrm{m}+3)=0
$
$
\mathrm{m}=5,-3
$
$
\text { C.F }=\mathrm{Ae}^{5 \mathrm{x}}+\mathrm{Be}^{-3 \mathrm{x}}
$

The general solution is $\mathrm{y}=\mathrm{Ae}^{5 \mathrm{x}}+\mathrm{Be}^{-3 \mathrm{x}}$
Differentiating (1) w.r.t ' $x$ '
$
\frac{d y}{d x}=5 \mathrm{~A} e^{5 x}-3 \mathrm{Be}^{-3 x}
$
Again differentiating (2) w.r.t $x$,
$
\frac{d^2 y}{d x^2}=25 \mathrm{Ae} e^{5 x}+9 \mathrm{Be}^{-3 x}
$
Using $x=0, \frac{d y}{d x}=0, \frac{d^2 y}{d x^2}=2$ in (2) and (3) we get,
$
\begin{aligned}
& 0=5 A-5 B \\
& 2=25 \mathrm{~A}+9 \mathrm{~B} \\
& \Rightarrow \quad 0=25 \mathrm{~A}-15 \mathrm{~B} \\
& 2=25 \mathrm{~A}+9 \mathrm{~B} \\
& -2=-24 B \\
& B=\frac{1}{12} \\
& 5 \mathrm{~A}=3 \mathrm{~B} \Rightarrow \mathrm{A}=\frac{3}{5}\left(\frac{1}{12}\right)=\frac{1}{20} \\
&
\end{aligned}
$
Using $\mathrm{A}=\frac{1}{20}$ and $\mathrm{B}=\frac{1}{12}$ in $(1)$, we get $y=\frac{e^{5 x}}{20}+\frac{e^{-3 x}}{12}$ as the solution
Question 6.
$
\left(4 \mathrm{D}^2+4 \mathrm{D}-3\right) y=\mathrm{e}^{2 x}
$
Solution:

The auxiliary equations is $4 \mathrm{~m}^2+4 \mathrm{~m}-3=0$
$
\begin{aligned}
(2 m+1)^2 & =4 \\
2 m+1 & =\pm 2 \Rightarrow m=\frac{1}{2}, \frac{-3}{2} \\
\text { C.F } & =\mathrm{A} e^{\frac{x}{2}}+\mathrm{B} e^{\frac{-3}{2} x} \\
\text { P.I. } & =\frac{1}{\phi(\mathrm{D})} f(x) \\
\text { P.I } & =\frac{1}{4 \mathrm{D}^2+4 \mathrm{D}-3} e^{2 x}
\end{aligned}
$
Replace D by 2
$
P . I=\frac{1}{4(2)^2+4(2)-3} e^{2 x}=\frac{1}{21} e^{2 x}
$
Hence the general solution is $y=\mathrm{C} . \mathrm{F}+$ P.I
$
\Rightarrow \quad y=\mathrm{A} e^{\frac{x}{2}}+\mathrm{B} e^{\frac{-3}{2} x}+\frac{e^{2 x}}{21}
$
Question 7.
$
\frac{d^2 y}{d x^2}+16 \mathrm{y}=0
$
Solution:
Given $\left(D^2+16\right) y=0$
The auxiliary equation is $\mathrm{m}^2+16=0$
$
\begin{aligned}
& \Rightarrow \mathrm{m}^2=-16 \\
& \Rightarrow \mathrm{m}=\pm 4 \mathrm{i}
\end{aligned}
$
It is of the form $\alpha \pm \mathrm{i} \beta, \alpha=0, \beta=4$
The complementary function (C.F) is $\mathrm{e}^{0 \mathrm{x}}[\mathrm{A} \cos 4 \mathrm{x}+\mathrm{B} \sin 4 \mathrm{x}$ ]
The general solution is $y=[A \cos 4 x+B \sin 4 x$ ]

Question 8.
$\left(D^2-3 D+2\right) y=e^{3 x}$ which shall vanish for $x=0$ and for $x=\log 2$
Solution:
$
\begin{aligned}
& \text { A.E is } m^2-3 m+2=0 \\
& \Rightarrow(m-2)(m-1)=0
\end{aligned}
$
$
\begin{aligned}
& \Rightarrow \mathrm{m}=2,1 \\
& \mathrm{CF}=\mathrm{Ae}^{2 \mathrm{x}}+\mathrm{Be}^{\mathrm{x}}
\end{aligned}
$
Replace D by 3
$
\text { P.I }=\frac{1}{\phi(\mathrm{D})} f(x)=\frac{1}{\mathrm{D}^2-3 \mathrm{D}+2}\left(e^{3 x}\right)
$
$
=\frac{1}{9-9+2}\left(e^{3 x}\right)=\frac{1}{2} e^{3 x}
$
The general solution is $\quad y=\mathrm{CF}+\mathrm{PI}$
$
y=\mathrm{A} e^{2 x}+\mathrm{Be}^x+\frac{e^{3 x}}{2}
$
Given that
$
\begin{array}{ll}
y=0, x=0 & \Rightarrow 0=\mathrm{A}+\mathrm{B}+\frac{1}{2} \\
y=0, x=\log 2 & \Rightarrow 0=4 \mathrm{~A}+2 \mathrm{~B}+4
\end{array}
$
Solving the two equations,
$
\begin{aligned}
4 \mathrm{~A}+2 \mathrm{~B}+4 & =0 \\
4 \mathrm{~A}+2 \mathrm{~B}+2 & =0 \\
\Rightarrow \quad 2 \mathrm{~B}-2 & =0 \Rightarrow \mathrm{B}=1, \mathrm{~A}=\frac{-3}{2}
\end{aligned}
$

So the solution is
$
y=\frac{-3}{2} e^{2 x}+e^x+\frac{e^{3 x}}{2}
$
Question 9.
$
\left(D^2+D-6\right) y=e^{3 x}+e^{-3 x}
$
Solution:
A.E is $\mathrm{m}^2+m-6=0$
$
(\mathrm{m}+3)(\mathrm{m}-2)=0
$
C.F is $\mathrm{A} e^{-3 x}+\mathrm{Be}^{2 x}$
Replace D by 3
$
m=-3,2
$
$
\mathrm{PI}_1=\frac{1}{\phi(\mathrm{D})} f(x)=\frac{1}{\mathrm{D}^2+\mathrm{D}-6} e^{3 x}
$
$
\begin{aligned}
& \mathrm{PI}_1=\frac{1}{9+3-6} e^{3 x}=\frac{e^{3 x}}{6} \\
& \mathrm{PI}_2=\frac{1}{\mathrm{D}^2+\mathrm{D}-6} e^{-3 x}
\end{aligned}
$
Replace D by $-3 \quad D^2+D-6=0$

Replace D by $-3$
$
\therefore \quad \mathrm{PI}_2=x \frac{1}{2 \mathrm{D}+1} e^{-3 x}
$
$
\mathrm{PI}_2=\frac{x e^{-3 x}}{-5}
$
The general solution is
$
\begin{aligned}
& y=\mathrm{CF}+\mathrm{PI}_1+\mathrm{PI}_2 \\
& y=\mathrm{A} e^{-3 x}+\mathrm{B} e^{2 x}+\frac{e^{3 x}}{6}-\frac{x e^{-3 x}}{5}
\end{aligned}
$
Question 10.
$
\left(\mathrm{D}^2-10 \mathrm{D}+25\right) \mathrm{y}=4 \mathrm{e}^{5 x}+5
$
Solution:
A.E is $\mathrm{m}^2-10 \mathrm{~m}+25=0$
$
\begin{aligned}
& \Rightarrow(\mathrm{m}-5)^2=0 \\
& \Rightarrow \mathrm{m}=5,5
\end{aligned}
$

C.F $=(A x+B) e^{5 x}$
$
\mathrm{PI}_1=\frac{1}{\mathrm{D}^2-10 \mathrm{D}+25}\left(4 e^{5 x}\right)=\frac{4 e^{5 x}}{(\mathrm{D}-5)^2}
$
Replace D by $5 \quad(\mathrm{D}-5)^2=0$
$
\begin{aligned}
& \therefore \quad \mathrm{PI}_1=4 x \frac{1}{2(\mathrm{D}-5)} e^{5 x} \\
& \mathrm{D}-5=0
\end{aligned}
$
Replace D by 5
$
\begin{aligned}
\mathrm{PI}_1 & =4 x^2\left(\frac{1}{2}\right) e^{5 x}=2 x^2 e^{5 x} \\
\mathrm{PI}_2 & =\frac{1}{(\mathrm{D}-5)^2} 5=\frac{1}{(\mathrm{D}-5)^2} 5 e^{o x}
\end{aligned}
$
Replace $\mathrm{D}=0 \Rightarrow \mathrm{PI}_2=\frac{1}{5}$
The general solution is $\quad y=\mathrm{CF}+\mathrm{PI}_1+\mathrm{PI}_2$
$
\Rightarrow \quad y=(\mathrm{A} x+\mathrm{B}) e^{5 x}+2 x^2 e^{5 x}+\frac{1}{5}
$
Question 11.
$
\left(4 D^2+16 \mathrm{D}+15\right) y=4 e^{\frac{-3}{2} x}
$
Solution:
A.E is $4 \mathrm{~m}^2+16 \mathrm{~m}+15=0 \quad(2 m+3)(2 m+5)=0$
$
m=\frac{-3}{2}, \frac{-5}{2}
$
C.F is $\mathrm{A} e^{\frac{-3}{2} x}+\mathrm{B} e^{\frac{-5}{2} x}$
$
\mathrm{PI}=\frac{1}{4 \mathrm{D}^2+16 \mathrm{D}+15} 4 e^{\frac{-3}{2} x}
$
Replace $\mathrm{D}$ by $\frac{-3}{2}, 4 \mathrm{D}^2+16 \mathrm{D}+15=0$
$
\begin{aligned}
\therefore \quad \mathrm{PI} & =4 x \frac{1}{(8 \mathrm{D}+16)} e^{\frac{-3}{2} x} \\
& =4 x\left(\frac{1}{4}\right) e^{\frac{-3}{2} x}\left(\text { Replace D by } \frac{-3}{2}\right)
\end{aligned}
$

$
=x e^{\frac{-3}{2} x}
$
The general solution is
$
\begin{aligned}
& y=\mathrm{CF}+\mathrm{PI} \\
& y=\mathrm{A} e^{\frac{-3}{2} x}+\mathrm{B} e^{\frac{-5}{2} x}+x e^{\frac{-3}{2} x}
\end{aligned}
$
Question 12.
$
\left(3 \mathrm{D}^2+\mathrm{D}-14\right) \mathrm{y}=13 \mathrm{e}^{2 \mathrm{x}}
$
Solution:
A.E is $3 m^2+m-14=0$
$
(3 m+7)(m-2)=0 \Rightarrow m=\frac{-7}{3}, 2
$
C.F is $\mathrm{A} e^{2 x}+\mathrm{B} e^{\frac{-7}{3} x}$
Replace D by 2
$
\mathrm{PI}=\frac{1}{3 \mathrm{D}^2+\mathrm{D}-14} 13 e^{2 x}
$
Replace D by $2 \quad 3 \mathrm{D}^2+\mathrm{D}-14=0$
$
\therefore \quad \text { PI }=x \frac{1}{6 \mathrm{D}+1} 13 e^{2 x}
$
The general solution is $\quad y=\mathrm{A} e^{2 x}+\mathrm{B} e^{\frac{-7}{3} x}+x e^{2 x}$
Question 13.
Suppose that the quantity demanded $\mathrm{Q}_{\mathrm{d}}=13-6 \mathrm{p}+2 \frac{d p}{d t}+\frac{d_2 p}{d t^2}$ and quantity supplied $\mathrm{Q}_{\mathrm{s}}$ $=-3+2 p$ where $p$ is the price. Find the equilibrium price for market clearance.
Solution:

For market clearance, the required condition is $\mathrm{Q}_{\mathrm{d}}=\mathrm{Q}_{\mathrm{s}}$
$
\begin{aligned}
& \Rightarrow 13-6 p+2 \frac{d p}{d t}+\frac{d^2 p}{d t^2}=-3+2 p \\
& \Rightarrow \quad \frac{d^2 p}{d t^2}+2 \frac{d p}{d t}-8 p=-16 \\
& \text { (or) } \quad\left(\mathrm{D}^2+2 \mathrm{D}-8\right) p=-16
\end{aligned}
$
The auxiliary equation is $m^2+2 m-8=0$
$
\begin{aligned}
(m+4)(m-2) & =0 \\
m & =-4,2 \\
\text { C.F } & =\mathrm{A} e^{-4 t}+\mathrm{B} e^{2 t} \\
\text { P.I }=\frac{1}{\phi(\mathrm{D})} f(t) & =\frac{1}{\mathrm{D}^2+2 \mathrm{D}-8}(-16) e^{0 t} \\
& =\frac{-16}{-8}(\text { Replace D by } 0) \\
& =2
\end{aligned}
$
The equilibrium price is $p=\mathrm{A} e^{-4 t}+\mathrm{B} e^{2 t}+2$