In the previous section, we saw how to check whether a function is the general solution or particular solution of a given differential equation. We saw some solved examples also. In this section, we will see a few more solved examples.
Solved example 25.7
(a) Solve the differential equation $\small{\frac{dy}{dt}~=~-t}$
Given that:
$\small{y=1}$ when $\small{t = 0}$.
(b) Solve the differential equation $\small{\frac{dy}{dt}~=~-t}$
Given that:
$\small{y=-1}$ when $\small{t = 0}$.
(c) Draw both results in the same graph
Solution:
Part (a):
1. We have: $\small{\frac{dy}{dt}~=~-t}$
•
Integrating both sides, we get:
$\small{y + \rm{C}_1~=~\frac{(-1)t^2}{2} + \rm{C}_2}$
$\small{\Rightarrow y~=~\frac{(-1)t^2}{2} + \rm{C}_2 - \rm{C}_1}$
$\small{\Rightarrow y~=~\frac{(-1)t^2}{2} + \rm{C}}$
•
This is the general solution.
2. Given that:
$\small{y=1}$ when $\small{t = 0}$.
•
Substituting in the general solution, we get:
$\small{1~=~\frac{(-1)(0)^2}{2} + \rm{C}}$
$\small{\Rightarrow 1~=~0 + \rm{C}}$
$\small{\Rightarrow 1~=~ \rm{C}}$
3. So the particular solution is:
$\small{y~=~\frac{(-1)t^2}{2} + 1}$
Part (b):
1. From part (a), we have:
$\small{y~=~\frac{(-1)t^2}{2} + \rm{C}}$
•
This is the general solution.
2. Given that:
$\small{y=-1}$ when $\small{t = 0}$.
•
Substituting in the general solution, we get:
$\small{-1~=~\frac{(-1)(0)^2}{2} + \rm{C}}$
$\small{\Rightarrow -1~=~0 + \rm{C}}$
$\small{\Rightarrow -1~=~ \rm{C}}$
3. So the particular solution is:
$\small{y~=~\frac{(-1)t^2}{2} - 1}$
Part (c):
The graphs are shown in fig.25.4 below:
 |
| Fig.25.4 |
•
The green curve represents the particular solution for part (a)
•
The red curve represents the particular solution for part (b)
Solved example 25.8
(a) Solve the differential equation $\small{\frac{dy}{dt}~=~2}$
Given that:
$\small{y=1}$ when $\small{t = 0}$.
(b) Solve the differential equation $\small{\frac{dy}{dt}~=~2}$
Given that:
$\small{y=-1}$ when $\small{t = 0}$.
(c) Draw both results in the same graph
Solution:
Part (a):
1. We have: $\small{\frac{dy}{dt}~=~2}$
• Integrating both sides, we get:
$\small{y + \rm{C}_1~=~2t + \rm{C}_2}$
$\small{\Rightarrow y~=~2t + \rm{C}_2 - \rm{C}_1}$
$\small{\Rightarrow y~=~2t + \rm{C}}$
• This is the general solution.
2. Given that:
$\small{y=1}$ when $\small{t = 0}$.
• Substituting in the general solution, we get:
$\small{1~=~2(0) + \rm{C}}$
$\small{\Rightarrow 1~=~0 + \rm{C}}$
$\small{\Rightarrow 1~=~ \rm{C}}$
3. So the particular solution is:
$\small{y~=~2t + 1}$
Part (b):
1. From part (a), we have:
$\small{y~=~2t + \rm{C}}$
• This is the general solution.
2. Given that:
$\small{y=-1}$ when $\small{t = 0}$.
• Substituting in the general solution, we get:
$\small{-1~=~2(0) + \rm{C}}$
$\small{\Rightarrow -1~=~0 + \rm{C}}$
$\small{\Rightarrow -1~=~ \rm{C}}$
3. So the particular solution is:
$\small{y~=~2t - 1}$
Part (c):
The graphs are shown in fig.25.5 below:
 |
| Fig.25.5 |
•
The green curve represents the particular solution for part (a)
•
The red curve represents the particular solution for part (b)
Solved example 25.9
(a) Solve the differential equation $\small{\frac{dv}{dt}~=~4t}$
Given that:
$\small{v=10}$ when $\small{t = 0}$.
(b) At what time does v increase to 100 or drop to 1?
Solution:
Part (a):
1. We have: $\small{\frac{dv}{dt}~=~4t}$
• Integrating both sides, we get:
$\small{v + \rm{C}_1~=~\frac{4t^2}{2} + \rm{C}_2}$
$\small{\Rightarrow v~=~2 t^2 + \rm{C}_2 - \rm{C}_1}$
$\small{\Rightarrow v~=~2t^2 + \rm{C}}$
• This is the general solution.
2. Given that:
$\small{v=10}$ when $\small{t = 0}$.
• Substituting in the general solution, we get:
$\small{10~=~2(0)^2 + \rm{C}}$
$\small{\Rightarrow 10~=~0 + \rm{C}}$
$\small{\Rightarrow 10~=~ \rm{C}}$
3. So the particular solution is:
$\small{v~=~2t^2 + 10}$
Part (b):
1. From part (a), we have:
$\small{v~=~2t^2 + 10}$
• This is the particular solution for the differential equation.
2. Substitute v = 100 in the particular solution:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{100} & {~=~} &{2 t^2~+~10} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{2t^2} & {~=~} &{90} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{t^2} & {~=~} &{45} \\
{~\color{magenta} 4 } &{{\Rightarrow}} &{t} & {~=~} &{3 \sqrt 5} \\
\end{array}}$
(We discard the -`ve root because, time cannot be -`ve.
3. We can write:
The velocity increase to 100 at the instant when the reading in the stop-watch is $\small{3 \sqrt 5}$ seconds.
An analysis of the above solved example 25.9, will give us greater insight into the basics of differential equations. The analysis can be written in 4 steps:
1. We are given the differential equation: $\small{\frac{dv}{dt}~=~4t}$
•
On the L.H.S, we have the derivative of velocity w.r.t time. So that derivative is the rate of change of velocity w.r.t time. In other words, it is the acceleration.
•
We can write: Acceleration of the given object is given by (4t)
2. If we integrate $\small{\frac{dv}{dt}}$, we will get the velocity v.
•
So if we integrate (4t), we will get the velocity.
•
That means, $\small{2t^2 + \rm{C}}$ is the expression for velocity.
•
This can be considered as the expression for velocity, for any object moving with an acceleration of (4t)
3. Our particular object has an initial value for velocity: When t = 0, v = 10.
•
Using this initial value, we obtained the useful result:
At any instant t, our particular object will be traveling with a velocity of $\small{2t^2 + 10}$
•
Note that, $\small{2t^2 + 10}$ is valid only for our particular object. For another object, even if acceleration is (4t), the initial value may be different and so the expression may be different.
4. We are asked to find the time when the velocity becomes 100 or 1.
•
For our particular object, the velocity can never become 1. The reason can be written in 3 steps:
(i) The acceleration is (4t), which is +ve. The initial velocity is 10.
(ii) A velocity of one, is less than a velocity of ten.
(iii) A lesser velocity can be achieved only if the object undergoes deceleration.
•
In other words, a lesser velocity can be achieved only if the object undergoes -`ve acceleration. Our particular object has a +ve acceleration.
Solved example 25.10
Verify that the function $\small{y~=~e^{-3x}}$ is a solution of the differential equation $\small{\frac{d^2y}{dx^2}~+~\frac{dy}{dx}~-~6y~=~0}$
Solution:
1. Find the derivatives from the given solution:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{y} & {~=~} &{e^{-3x}} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{{\frac{dy}{dx}}} & {~=~} &{(-3)e^{-3x}} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{{\frac{d^2y}{dx^2}}} & {~=~} &{(-3)(-3)e^{-3x}~=~9 e^{-3x}} \\
\end{array}}$
2. Substitute the above derivatives in the given differential equation:
• The given differential equation is: $\small{\frac{d^2y}{dx^2}~+~\frac{dy}{dx}~-~6y~=~0}$
Substituting, we get: $\small{9 e^{-3x}~+~(-3) e^{-3x}~-~6e^{-3x}~=~0}$
Which is true.
• So $\small{y = e^{-3x}}$ is indeed a solution.
Solved example 25.11
Verify that the function $\small{y~=~a \cos x~+~b \sin x}$, where a, b ∈ R is a solution of the
differential equation
$\small{\frac{d^2y}{dx^2}~+~y~=~0}$
Solution:
1. Find the derivatives from the given solution:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{y} & {~=~} &{a \cos x~+~b \sin x} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{{\frac{dy}{dx}}} & {~=~} &{-a \sin x~+~b \cos x} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{{\frac{d^2y}{dx^2}}} & {~=~} &{-a \cos x~-~b \sin x} \\
\end{array}}$
2. Substitute the above derivatives in the given differential equation:
• The given differential equation is: $\small{\frac{d^2y}{dx^2}~+~y~=~0}$
Substituting, we get: $\small{\left[-a \cos x~-~b \sin x \right]~+~\left[a \cos x~+~b \sin x \right]~=~0}$
$\small{\Rightarrow a(\cos x~-~\cos x)~+~b(\sin x~-~\sin x)~=~0}$
Which is true.
• So $\small{y = a \cos x~+~b \sin x}$ is indeed a solution.
Solved example 25.12
Verify that the implicit function $\small{y~-~\cos y~=~x}$, is a solution of the
differential equation
$\small{(y \sin y ~+~\cos y~+~x)\frac{dy}{dx}~=~y}$
Solution:
1. We already know the method to find $\small{\frac{dy}{dx}}$ when we are given an implicit function. (see section 21.8)
• So in out present case, we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{y~-~\cos y} & {~=~} &{x} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{{\frac{dy}{dx}~-~\left((-\sin y)\frac{dy}{dx} \right)}} & {~=~} &{1} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{{\frac{dy}{dx}~+~\left(\sin y \right)\frac{dy}{dx}}} & {~=~} &{1} \\
{~\color{magenta} 4 } &{{\Rightarrow}} &{{\left(1~+~\sin y \right)\frac{dy}{dx}}} & {~=~} &{1} \\
{~\color{magenta} 5 } &{{\Rightarrow}} &{{\frac{dy}{dx}}} & {~=~} &{\frac{1}{1~+~\sin y}} \\
\end{array}}$
2. Substitute the above derivative in the given differential equation:
• The given differential equation is: $\small{(y \sin y ~+~\cos y~+~x)\frac{dy}{dx}~=~y}$
Substituting, we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{(y \sin y ~+~\cos y~+~x)\frac{1}{1~+~\sin y}} & {~=~} &{y} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{{(y \sin y ~+~\cos y~+~y~-~\cos y)\frac{1}{1~+~\sin y}}} & {~=~} &{y} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{{(y \sin y ~+~y)\frac{1}{1~+~\sin y}}} & {~=~} &{y} \\
{~\color{magenta} 4 } &{{\Rightarrow}} &{{y(\sin y ~+~1)\frac{1}{1~+~\sin y}}} & {~=~} &{y} \\
\end{array}}$
Which is true.
• So the implicit function $\small{y-\cos y~=~x}$ is indeed a solution.
Solved example 25.13
Verify that the implicit function $\small{x~+~y~=~\tan^{-1} y}$, is a solution of the
differential equation
$\small{y^2\frac{dy}{dx}~+~y^2~+~1~=~0}$
Solution:
1. We already know the method to find $\small{\frac{dy}{dx}}$ when we are given an implicit function. (see section 21.8)
• So in out present case, we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{x~+~y} & {~=~} &{\tan^{-1}y} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{{1~+~\frac{dy}{dx}}} & {~=~} &{\frac{1}{1 + y^2}\frac{dy}{dx}} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{\frac{1}{1 + y^2}\frac{dy}{dx}~-~\frac{dy}{dx}} & {~=~} &{1} \\
{~\color{magenta} 4 } &{{\Rightarrow}} &{\frac{dy}{dx}\left(\frac{1}{1 + y^2}~-~1 \right)} & {~=~} &{1} \\
{~\color{magenta} 5 } &{{\Rightarrow}} &{\frac{dy}{dx}\left(\frac{1-1 - y^2}{1 + y^2} \right)} & {~=~} &{1} \\
{~\color{magenta} 6 } &{{\Rightarrow}} &{\frac{dy}{dx}\left(\frac{- y^2}{1 + y^2} \right)} & {~=~} &{1} \\
{~\color{magenta} 7 } &{{\Rightarrow}} &{\frac{dy}{dx}} & {~=~} &{\frac{-1- y^2}{y^2}} \\
\end{array}}$
2. Substitute the above derivative in the given differential equation:
• The given differential equation is: $\small{y^2\frac{dy}{dx}~+~y^2~+~1~=~0}$
Substituting, we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{y^2\left(\frac{-1- y^2}{y^2} \right)~+~y^2~+~1} & {~=~} &{0} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{-1- y^2~+~y^2~+~1} & {~=~} &{0} \\
\end{array}}$
Which is true.
• So the implicit function $\bf{x~+~y~=~\tan^{-1} y}$ is indeed a solution.
In the next section, we will see order and degree of differential equations.
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