In the previous section,
we completed a discussion on application of integrals. In
this chapter, we will see differential equations.
Some basic details can be written in 8 steps:
1. Consider a car moving with a constant acceleration 'a'. We know that, the distance traveled by the car is given by the equation: $\small{s = ut + \frac{1}{2}at^2}$
Where,
s = Distance traveled in time t
u = the initial velocity (a constant)
a = acceleration with which the car is traveling (a constant)
•
So distance depends on t. In other words, distance is a function of t.
♦ s is the dependent variable.
♦ t is the independent variable.
•
We can write: $\small{s = f(t) = \text{u}t + \frac{1}{2}\text{a}t^2}$
2. We know that, velocity is the derivative of distance w.r.t time. So we can write:
$\small{v = f'(t) = \frac{ds}{dt}=\text{u} + \frac{1}{2}\text{a}(2t)}$
$\small{\Rightarrow v = f'(t) = \frac{ds}{dt}=\text{u} + \text{a}t}$
$\small{\Rightarrow v =\text{u} + \text{a}t}$
3. We also know that, acceleration is the derivative of velocity w.r.t time. So we can write:
$\small{\frac{dv}{dt} = f''(t) = \frac{d^2s}{dt^2}=0 + \text{a}}$
$\small{\Rightarrow \frac{dv}{dt} = f''(t) = \frac{d^2s}{dt^2}=\text{a}}$
$\small{\Rightarrow \frac{dv}{dt}=\text{a}}$
4. Let us now think in a reverse manner. If we are given the acceleration, can we find the velocity?
•
Note that, the car is moving with a constant acceleration 'a'. So the velocity is continuously changing.
•
Then the question is:
If we are given the acceleration 'a', can we write an expression for the velocity?
(If we can write such an expression, we will be able to find the velocity at any instant t)
•
Let us try to write such an expression:
In step (3), we wrote: $\small{\frac{dv}{dt}=\text{a}}$
Integrating both sides, we get:
$\small{v + \rm{C}_1=\text{a}t + \rm{C}_2}$
$\small{\Rightarrow v =\text{a}t + \rm{C}_2 - \rm{C}_1}$
$\small{\Rightarrow v =\text{a}t + \rm{C}}$
5. The expression that we obtained in the above step (4), is applicable to any object moving with a constant acceleration 'a'.
•
It is called the general solution of the differential equation $\small{\frac{dv}{dt}=\text{a}}$.
6. To put the above general equation to practical use, we need to find the value of the constant 'C'. For that, we must have some additional information.
•
Let, at the instant when the stop-watch is turned on, the car be moving at a velocity of 9 m/s
•
At the time when the stop-watch is turned on, time t = 0.
•
Substituting these values in the general solution, we get:
$\small{9 =\text{a}(0) + \rm{C}}$
$\small{\Rightarrow 9 =\rm{C}}$
•
So for the car, we can write the expression for velocity:
$\small{v =\text{a}t + 9}$
•
This is called particular solution of the differential equation $\small{\frac{dv}{dt}=\text{a}}$.
(Note that, this particular solution is same as our familiar equation $\small{v =\text{u} + \text{a}t}$ where u is the initial velocity)
•
We were able to write the particular solution by using the fact that:
Initially, when t = 0, velocity is 9 m/s
7. The following graph in fig.25.1 will help us to understand the difference between general solution and particular solution. It is assumed that, the constant a = 3/4 ms−2.
 |
| Fig.25.1 |
•
The topmost red line represents $\small{v =\frac{3}{4} t + 14}$, where the constant C = 14
•
The second red line from top, represents $\small{v =\frac{3}{4} t + 11}$, where the constant C = 11
•
In this way, we can draw infinite number of parallel red lines.
•
But we are interested in the green line which represents the unique particular solution, where C = 9
•
Note that, the graph of the particular solution, passes through (0,9).
• (0,9) is an ordered pair.
♦ The x-coordinate (abscissa) gives time
♦ The y-coordinate (ordinate) gives velocity
It is a velocity-time graph.
8. From the above steps, we learned six new items:
(i) We saw the differential equation $\small{\frac{dv}{dt}=\text{a}}$
(ii) In step (4), we got the general solution of that differential equation. We got:
$\small{v =\text{a}t + \rm{C}}$
(iii) In step (6), we got the particular solution of that differential equation. We got:
$\small{v =\text{a}t + 9}$
(iv) Differential equation is a relation between:
♦ Derivative of the dependent variable v
♦ and
♦ The independent variable t
(v) Solution (general or particular) of the differential equation is a relation between:
♦ Dependent variable v
♦ and
♦ The independent variable t
(vi) In the graph of step(7), we saw the difference between general solution and particular solution.
Let us see another example. It can be written in 7 steps:
1. Consider the same car that we saw in the example above. It is moving with a constant acceleration 'a'.
2. We saw that,
$\small{v = f'(t) = \frac{ds}{dt}=\text{u} + \text{a}t}$
3. Let us now think in a reverse manner. If we are given the velocity, can we find the distance traveled?
• Note that, the car is moving with a constant acceleration 'a'. So the velocity is continuously changing. Consequently, the distance traveled in unit time is also continuously changing.
• Then the question is:
If we are given the expression for velocity, can we write an expression for the distance traveled?
(If we can write such an expression, we will be able to find the distance traveled any instant t)
• Let us try to write such an expression:
In step (2), we wrote: $\small{\frac{ds}{dt}=\text{u} + \text{a}t}$ Integrating both sides, we get:
$\small{s + \rm{C}_1=\text{u}t + \text{a} \frac{t^2}{2}+ \rm{C}_2}$
$\small{\Rightarrow s =\text{u}t + \frac{1}{2} \text{a} t^2 + \rm{C}_2 - \rm{C}_1}$
$\small{\Rightarrow s =\text{u}t + \frac{1}{2} \text{a} t^2 + \rm{C}}$
4. The expression that we obtained in the above step (3), is applicable to any object moving with a constant acceleration 'a' and initial velocity u.
• It is called the general solution of the differential equation $\small{\frac{ds}{dt}=\text{u} + \text{a}t}$
5. To put the above general equation to practical use, we need to find the value of the constant 'C'. For that, we must have some additional information.
• At the instant when the stop-watch is turned on, time t = 0.
• In a duration of zero seconds, the distance 's' traveled will be zero.
• Substituting these values in the general solution, we get:
$\small{0 =\text{u}(0) + \frac{1}{2} \text{a} (0)^2 + \rm{C}}$
$\small{\Rightarrow 0 = 0 + 0 + \rm{C}}$
$\small{\Rightarrow 0 =\rm{C}}$
• So for the car, we can write the expression for distance:
$\small{s =\text{u}t + \frac{1}{2} \text{a} t^2 + 0}$
$\small{\Rightarrow s =\text{u}t + \frac{1}{2} \text{a} t^2}$
• This is called particular solution of the differential equation $\small{\frac{ds}{dt}=\text{u} + \text{a}t}$.
(Note that, this particular solution is same as our familiar equation $\small{s =\text{u} t +\frac{1}{2} \text{a} t^2}$ where u is the initial velocity)
• We were able to write the particular solution by using the fact that:
Initially, when t = 0, distance traveled is zero.
6. The following graph in fig.25.2 will help us to understand the difference between general solution and particular solution. It is assumed that,
♦ the constant a = 3/4 ms−2.
♦ the constant u = 9 ms−1.
 |
| Fig.25.2 |
• The topmost red curve represents $\small{s =\text{u}t + \frac{1}{2} \text{a} t^2 + 19}$ , where the constant C = 19
• The second red curve from top, represents $\small{s =\text{u}t + \frac{1}{2} \text{a} t^2 + 15}$ , where the constant C = 15
• In this way, we can draw infinite number of red curves.
• But we are interested in the green curve which represents the unique particular solution, where C = 0
• Note that, the graph of the particular solution, passes through (0,0).
• (0,0) is an ordered pair.
♦ The x-coordinate (abscissa) gives time
♦ The y-coordinate (ordinate) gives distance
It is a distance-time graph.
7. From the above steps, we learned four new items:
(i) We saw a differential equation $\small{\frac{ds}{dt}=\text{u} + \text{a}t}$
(ii) In step (3), we got the general solution of that differential equation. We got:
$\small{s =\text{u}t + \frac{1}{2} \text{a} t^2 + \rm{C}}$
(iii) In step (5), we got the particular solution of that differential equation. We got:
$\small{s =\text{u}t + \frac{1}{2} \text{a} t^2}$
(iv) Differential equation is a relation between:
♦ Derivative of the dependent variable s
♦ and
♦ The independent variable t
(v) Solution (general or particular) of the differential equation is a relation between:
♦ Dependent variable s
♦ and
♦ The independent variable t
• In the graph of step(6), we saw the difference between general solution and particular solution.
Now we will see an important feature of the general solution. It can be written in two steps:
1. A solution must satisfy the given equation. For example:
If someone says that x = 8 is a solution of the equation x2 − 6x −16 = 0, we can cross check by substituting x = 8.
•
We get: 82 − 6(8) − 16 = 0
⇒ 64 − 48 − 16 = 0
⇒ 16 − 16 = 0, which is true.
•
So x = 8 is indeed a solution of the equation x2 − 6x −16 = 0
2. In our present case, we saw how the general solution of a differential equation can be obtained. After obtaining the general solution, we must cross check. (In some cases, we will be asked to cross check a given solution) The checking can be done in 4 steps.
◼ Consider the first example.
(i) We obtained $\small{v =\text{a}t + \rm{C}}$ as the general solution for the differential equation $\small{\frac{dv}{dt}=\text{a}}$.
(ii) Differentiating the general solution, we get: $\small{\frac{dv}{dt} = \text {a}}$
(iii) Substituting this in the given differential equation, we get: $\small{\text{a}=\text{a}}$, which is true.
•
So $\small{v =\text{a}t + \rm{C}}$ is indeed the general solution of the given differential equation.
(iv) If particular solution is available, then it must also be cross checked.
•
In the first example, the particular solution is: $\small{v =\text{a}t + 9}$.
•
We have the additional information that, when t = 0, velocity is 9. In other words, (0,9) is a point in the solution.
•
Substituting this point in the particular solution, we get: $\small{9 =\text{a}(0) + 9}$
$\small{\Rightarrow 9 = 0 + 9}$, which is true.
•
So $\small{v =\text{a}t + 9}$ is indeed the particular solution of the given differential equation.
◼ Consider the second example.
(i) We obtained $\small{s =\text{u}t + \frac{1}{2} \text{a} t^2 + \rm{C}}$ as the general solution for the differential equation $\small{\frac{ds}{dt}=\text{u} + \text{a}t}$.
(ii) Differentiating the general solution, we get: $\small{\frac{ds}{dt} = \text {u} + \frac{1}{2} \text{a} (2t) + 0}$
$\small{\Rightarrow \frac{ds}{dt} = \text {u} + \text{a}t}$
(iii) Substituting this in the given differential equation, we get:
$\small{\text {u} + \text{a}t=\text{u} + \text{a}t}$, which is true.
•
So $\small{s =\text{u}t + \frac{1}{2} \text{a} t^2 + \rm{C}}$ is indeed the general solution of the given differential equation.
(iv) If particular solution is available, then it must also be cross checked.
•
In the second example, the particular solution is: $\small{s =\text{u}t + \frac{1}{2} \text{a} t^2}$.
•
We have the additional information that, when t = 0, distance is zero. In other words, (0,0) is a point in the solution.
Substituting this point in the particular solution, we get: $\small{0 =\text{u}(0) + \frac{1}{2} \text{a} (0)^2}$
$\small{\Rightarrow 0 = 0 + 0}$, which is true.
•
So $\small{s =\text{u}t + \frac{1}{2} \text{a} t^2}$ is indeed the particular solution of the given differential equation.
Now we will see some solved examples:
Solved example 25.1
Check whether the function $\small{y = \frac{4x^3}{3} + \rm{C}}$ is the general solution of the differential equation $\small{\frac{dy}{dx} = 4x^2}$
If $\small{y=-30}$ when $\small{x = -3}$, then find the particular solution.
Solution:
Part I: Checking whether the given function is the general solution.
1. Find the derivative $\small{\frac{dy}{dx}}$ from the given general solution:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{y} & {~=~} &{\frac{4x^3}{3} + \rm{C}} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{{\frac{dy}{dx}}} & {~=~} &{\frac{4}{3} \left(3x^2 \right)+0} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{4x^2} \\
\end{array}}$
2. Substitute the above derivative in the given differential equation:
• The given differential equation is: $\small{\frac{dy}{dx} = 4x^2}$
Substituting, we get: $\small{4x^2 = 4x^2}$
Which is true.
•
So $\small{y = \frac{4x^3}{3} + \rm{C}}$ is indeed the general solution.
Part II: Finding the particular solution.
1. The general solution contains the constant C. We must find the value of C.
•
Given that, $\small{y=-30}$ when $\small{x = -3}$
•
Substituting these in the general solution, we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{y} & {~=~} &{\frac{4x^3}{3} + \rm{C}} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{-30} & {~=~} &{\frac{4(-3)^3}{3} + \rm{C}} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{-30} & {~=~} &{4(-1)3^2 + \rm{C}} \\
{~\color{magenta} 4 } &{{\Rightarrow}} &{-30 + 36} & {~=~} &{\rm{C}} \\
{~\color{magenta} 5 } &{{\Rightarrow}} &{6} & {~=~} &{\rm{C}} \\
\end{array}}$
2. So the particular solution is:
$\small{y = \frac{4x^3}{3} + 6}$
Solved example 25.2
Check whether the
function $\small{y = \frac{3x^4}{4} + \rm{C}}$ is the general solution
of the differential equation $\small{\frac{dy}{dx} = 3x^3}$
If $\small{y=\frac{19}{4}}$ when $\small{x = 1}$, then find the particular solution.
Solution:
Part I: Checking whether the given function is the general solution.
1. Find the derivative $\small{\frac{dy}{dx}}$ from the given general solution:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{y} & {~=~} &{\frac{3x^4}{4} + \rm{C}} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{{\frac{dy}{dx}}} & {~=~} &{\frac{3}{4} \left(4x^3 \right)+0} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{3x^3} \\
\end{array}}$
2. Substitute the above derivative in the given differential equation:
• The given differential equation is: $\small{\frac{dy}{dx} = 3x^3}$
Substituting, we get: $\small{3x^3 = 3x^3}$
Which is true.
• So $\small{y = \frac{3x^4}{4} + \rm{C}}$ is indeed the general solution.
Part II: Finding the particular solution.
1. The general solution contains the constant C. We must find the value of C.
• Given that, $\small{y=\frac{19}{4}}$ when $\small{x = 1}$
• Substituting these in the general solution, we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{y} & {~=~} &{\frac{3x^4}{4} + \rm{C}} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{\frac{19}{4}} & {~=~} &{\frac{3(1)^4}{4} + \rm{C}} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{\frac{19}{4}} & {~=~} &{\frac{3}{4} + \rm{C}} \\
{~\color{magenta} 4 } &{{\Rightarrow}} &{\frac{16}{4}} & {~=~} &{\rm{C}} \\
{~\color{magenta} 5 } &{{\Rightarrow}} &{4} & {~=~} &{\rm{C}} \\
\end{array}}$
2. So the particular solution is:
$\small{y = \frac{3x^4}{4} + 4}$
Solved example 25.3
Check whether the function $\small{y = \rm{C}\,e^{x^2}}$ is the general solution of the differential equation $\small{\frac{dy}{dx} = 2xy}$
If $\small{y=\frac{1}{2}}$ when $\small{x = 0}$, then find the particular solution.
Solution:
Part I: Checking whether the given function is the general solution.
1. Find the derivative $\small{\frac{dy}{dx}}$ from the given general solution:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{y} & {~=~} &{\rm{C}\,e^{x^2}} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{{\frac{dy}{dx}}} & {~=~} &{\rm{C}\,e^{x^2}(2x)} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{2\rm{C}\,xe^{x^2}} \\
\end{array}}$
2. Substitute the above derivative and y in the given differential equation:
• The given differential equation is: $\small{\frac{dy}{dx} = 3xy}$
Substituting, we get: $\small{2\rm{C}\,xe^{x^2} = 2x(\rm{C}\,e^{x^2})}$
$\small{\Rightarrow 2\,xe^{x^2} = 2x(e^{x^2})}$
Which is true.
• So $\small{y = \rm{C}\,e^{x^2}}$ is indeed the general solution.
Part II: Finding the particular solution.
1. The general solution contains the constant C. We must find the value of C.
• Given that, $\small{y=\frac{1}{2}}$ when $\small{x = 0}$
• Substituting these in the general solution, we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{y} & {~=~} &{\rm{C}\,e^{x^2}} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{\frac{1}{2}} & {~=~} &{\rm{C}\,e^{(0)^2}} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{\frac{1}{2}} & {~=~} &{\rm{C}\,e^{0}} \\
{~\color{magenta} 4 } &{{\Rightarrow}} &{\frac{1}{2}} & {~=~} &{\rm{C}(1)} \\
{~\color{magenta} 5 } &{{\Rightarrow}} &{\frac{1}{2}} & {~=~} &{\rm{C}} \\
\end{array}}$
2. So the particular solution is:
$\small{y = \frac{1}{2} e^{x^2}}$
3. The graph is shown in fig.25.3 below:
 |
| Fig.25.3 |
•
The top most red curve represents the solution when C = 2
•
The second red curve from top represents the solution when C = 1
•
The bottom most red curve represents the solution when C = 1/3
•
In this manner, infinite number of red curves can be drawn. But there is only one green curve, which represents the particular solution. For that green curve, C = 1/2.
Solved example 25.4
Check whether the function $\small{y = \rm{C}\,e^{-1/x}}$ is the
general solution of the differential equation $\small{\frac{dy}{dx} x^2 = y}$
If $\small{y=\frac{2}{e}}$ when $\small{x = 1}$, then find the particular solution.
Solution:
Part I: Checking whether the given function is the general solution.
1. Find the derivative $\small{\frac{dy}{dx}}$ from the given general solution:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{y} & {~=~} &{\rm{C}\,e^{-1/x}} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{{\frac{dy}{dx}}} & {~=~} &{\rm{C}\,e^{-1/x}\left(\frac{1}{x^2} \right)} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{\frac{\rm{C}\,e^{-1/x}}{x^2} } \\
\end{array}}$
2. Substitute the above derivative and y in the given differential equation:
• The given differential equation is: $\small{\frac{dy}{dx} x^2 = y}$
Substituting, we get: $\small{\left(\frac{\rm{C}\,e^{-1/x}}{x^2} \right) x^2 = \rm{C}\,e^{-1/x}}$
Which is true.
• So $\small{y = \rm{C}\,e^{-1/x}}$ is indeed the general solution.
Part II: Finding the particular solution.
1. The general solution contains the constant C. We must find the value of C.
• Given that, $\small{y=\frac{2}{e}}$ when $\small{x = 1}$
• Substituting these in the general solution, we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{y} & {~=~} &{\rm{C}\,e^{-1/x}} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{\frac{2}{e}} & {~=~} &{\rm{C}\,e^{-1/1}} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{\frac{2}{e}} & {~=~} &{\rm{C}\left(\frac{1}{e} \right)} \\
{~\color{magenta} 4 } &{{\Rightarrow}} &{\frac{2}{e}} & {~=~} &{\frac{\rm{C}}{e}} \\
{~\color{magenta} 5 } &{{\Rightarrow}} &{2} & {~=~} &{\rm{C}} \\
\end{array}}$
2. So the particular solution is:
$\small{y = 2 e^{-1/x}}$
Solved example 25.5
Check whether the function $\small{u = \sin^{-1}\left(e^{\rm{C} + t} \right)}$ is the general solution of the differential equation $\small{\frac{du}{dt} = \tan u}$
If $\small{u=\frac{\pi}{2}}$ when $\small{t = 1}$, then find the particular solution.
Solution:
Part I: Checking whether the given function is the general solution.
1. Find the derivative $\small{\frac{du}{dt}}$ from the given general solution:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{u} & {~=~} &{\sin^{-1}\left(e^{\rm{C} + t} \right)} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{{\frac{du}{dt}}} & {~=~} &{\frac{e^{\rm{C} + t}}{\sqrt{1 - \left(e^{\rm{C} + t} \right)^2 }}} \\
\end{array}}$
2. Substitute the above derivative and u in the given differential equation:
• The given differential equation is: $\small{\frac{du}{dt} = \tan u}$
Substituting, we get: $\small{\frac{e^{\rm{C} + t}}{\sqrt{1 - \left(e^{\rm{C} + t} \right)^2 }} = \tan\left[\sin^{-1}\left(e^{\rm{C} + t} \right) \right]}$
3. So our next task is to find: $\small{ \tan\left[\sin^{-1}\left(e^{\rm{C} + t} \right) \right]}$
Let $\small{y = \sin^{-1}\left(e^{\rm{C} + t} \right)}$. Then $\small{\sin y = e^{\rm{C} + t}}$
we want tan y.
• Since $\small{\sin y = e^{\rm{C} + t}}$, we can write:
$\small{\cos y = \sqrt{1~-~\left(e^{\rm{C} + t} \right)^2}}$
• Then $\small{\tan y~=~\frac{e^{\rm{C} + t}}{\sqrt{1~-~\left(e^{\rm{C} + t} \right)^2}}}$
4. So from step (2), we get:
$\small{\frac{e^{\rm{C} + t}}{\sqrt{1 - \left(e^{\rm{C} + t} \right)^2 }} = \tan\left[\sin^{-1}\left(e^{\rm{C} + t} \right) \right]~=~\frac{e^{\rm{C} + t}}{\sqrt{1~-~\left(e^{\rm{C} + t} \right)^2}}}$
Which is true.
• So $\small{u = \sin^{-1}\left(e^{\rm{C} + t} \right)}$ is indeed the general solution.
Part II: Finding the particular solution.
1. The general solution contains the constant C. We must find the value of C.
• Given that, $\small{u=\frac{\pi}{2}}$ when $\small{t = 1}$
• Substituting these in the general solution, we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{u} & {~=~} &{\sin^{-1}\left(e^{\rm{C}+t} \right)} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{\frac{\pi}{2}} & {~=~} &{\sin^{-1}\left(e^{\rm{C}+1} \right)} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{\sin \left(\frac{\pi}{2} \right)} & {~=~} &{e^{\rm{C}+1} } \\
{~\color{magenta} 4 } &{{\Rightarrow}} &{1} & {~=~} &{e(e^{\rm{C}})} \\
{~\color{magenta} 5 } &{{\Rightarrow}} &{e^{\rm{C}}} & {~=~} &{\frac{1}{e}~=~e^{-1}} \\
{~\color{magenta} 6 } &{{\Rightarrow}} &{\rm{C}} & {~=~} &{-1} \\
\end{array}}$
2. So the particular solution is:
$\small{u = \sin^{-1}\left(e^{-1+t} \right)}$
Solved example 25.6
Check whether the function $\small{x = \rm{C}~-~\frac{1}{8} \sin(2t)~-~\frac{1}{32} \sin(4t)}$ is the general solution of the differential equation
$\small{8 \frac{dx}{dt} = -2 \cos (2t)~-~\cos(4t)}$
If $\small{x=\pi}$ when $\small{t = \pi}$, then find the particular solution.
Solution:
Part I: Checking whether the given function is the general solution.
1. Find the derivative $\small{\frac{dx}{dt}}$ from the given general solution:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{x} & {~=~} &{\rm{C}~-~\frac{1}{8} \sin(2t)~-~\frac{1}{32} \sin(4t)} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{{\frac{dx}{dt}}} & {~=~} &{0~-~\frac{1}{8} (2)\cos(2t)~-~\frac{1}{32} (4)\cos(4t)} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{{\frac{dx}{dt}}} & {~=~} &{-\frac{1}{4}\cos(2t)~-~\frac{1}{8} \cos(4t)} \\
\end{array}}$
2. Substitute the above derivative and x in the given differential equation:
• The given differential equation is: $\small{8 \frac{dx}{dt} = -2 \cos (2t)~-~\cos(4t)}$
Substituting, we get:
$\small{8 \left[-\frac{1}{4}\cos(2t)~-~\frac{1}{8} \cos(4t) \right] = -2 \cos (2t)~-~\cos(4t)}$
$\small{\Rightarrow \left[-2 \cos(2t)~-~ \cos(4t) \right] = -2 \cos (2t)~-~\cos(4t)}$
Which is true.
• So $\small{x = \rm{C}~-~\frac{1}{8} \sin(2t)~-~\frac{1}{32} \sin(4t)}$ is indeed the general solution.
Part II: Finding the particular solution.
1. The general solution contains the constant C. We must find the value of C.
• Given that, $\small{x=\pi}$ when $\small{t = \pi}$
• Substituting these in the general solution, we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{x} & {~=~} &{\rm{C}~-~\frac{1}{8} \sin(2t)~-~\frac{1}{32} \sin(4t)} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{\pi} & {~=~} &{\rm{C}~-~\frac{1}{8} \sin(2\pi)~-~\frac{1}{32} \sin(4 \pi)} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{\pi} & {~=~} &{\rm{C}~-~\frac{1}{8} (0)~-~\frac{1}{32} (0)} \\
{~\color{magenta} 4 } &{{\Rightarrow}} &{\pi} & {~=~} &{\rm{C}~-~0} \\
{~\color{magenta} 5 } &{{\Rightarrow}} &{\rm{C}} & {~=~} &{\pi} \\
\end{array}}$
2. So the particular solution is:
$\small{x ~=~ \pi~-~\frac{1}{8} \sin(2t)~-~\frac{1}{32} \sin(4t)}$
In the next section, we will see a few more solved examples.
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