In the previous section, we saw formation of a differential equation whose general solution is given. We saw a solved example also. In this section, we will see a few more solved examples.
Solved example 25.17
The equation $y^2~=~a(b^2 - x^2)$ gives a family of curves.
Form a differential equation representing the above family of curves, by eliminating arbitrary constants a and b.
Solution:
1. To eliminate 'a' and 'b', we differentiate the equation as shown below:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{y^2} & {~=~} &{a(b^2 - x^2)} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{2y\,y'} & {~=~} &{0~-~2x\,a} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{y\,y'} & {~=~} &{-ax} \\
{~\color{magenta} 4 } &{{\Rightarrow}} &{y\,y''~+~y'\,y'} & {~=~} &{-a} \\
{~\color{magenta} 5 } &{{\Rightarrow}} &{y\,y''~+~(y')^2} & {~=~} &{\frac{y\,y'}{x}} \\
{~\color{magenta} 6 } &{{\Rightarrow}} &{xy\,y''~+~x\,(y')^2~-~y\,y'} & {~=~} &{0} \\
\end{array}}$
◼ Remarks:
In [(5) magenta color], we use the result from [(3) magenta color] to eliminate 'a'
2. We got a differential equation without arbitrary constants.
• We can write:
the differential equation: $\small{xy\,y''~+~x\,(y')^2~-~y\,y'~=~0}$ represents the given family of curves.
Solved example 25.18
The equation $y~=~a \sin(x+b)$ gives a family of curves.
Form a differential equation representing the above family of curves, by eliminating arbitrary constants a and b.
Solution:
1. To eliminate 'a' and 'b', we differentiate the equation as shown below:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{y} & {~=~} &{a \sin(x+b)} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{y'} & {~=~} &{a \cos (x+b)} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{y''} & {~=~} &{-a \sin(x+b)} \\
{~\color{magenta} 4 } &{{\Rightarrow}} &{y''} & {~=~} &{-y} \\
{~\color{magenta} 5 } &{{\Rightarrow}} &{y''~+~y} & {~=~} &{0} \\
\end{array}}$
◼ Remarks:
In [(4) magenta color], we use the original equation from [(1) magenta color] to eliminate 'a' and 'b'
2. We got a differential equation without arbitrary constants.
• We can write:
the differential equation: $\small{y''~+~y~=~0}$ represents the given family of curves.
Solved example 25.19
Form the differential equation representing the family of ellipses having foci on the x-axis and center at the origin.
Solution:
1. We know that, the family mentioned in the question is given by the equation: $\small{\frac{x^2}{a^2}~+~\frac{y^2}{b^2}~=~1}$ (see section 11.4)
A few members of the family are shown in the fig.25.12 below:
 |
| Fig.25.12 |
For the green ellipse, a = 5 and b = 3
For the red ellipse, a = 3 and b = 2
For the yellow ellipse, a = 7 and b = 4
2. To eliminate 'a' and 'b', we differentiate the equation as shown below:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\frac{x^2}{a^2}~+~\frac{y^2}{b^2}} & {~=~} &{1} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{\frac{2x}{a^2}~+~\frac{2y\,y'}{b^2}} & {~=~} &{0} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{\frac{x}{a^2}~+~\frac{y\,y'}{b^2}} & {~=~} &{0} \\
{~\color{magenta} 4 } &{{\Rightarrow}} &{x b^2~+~y\,y'\,a^2} & {~=~} &{0} \\
{~\color{magenta} 5 } &{{\Rightarrow}} &{ b^2~+~a^2\left[y\,y''~+~(y')^2 \right]} & {~=~} &{0} \\
{~\color{magenta} 6 } &{{\Rightarrow}} &{ \frac{(-1)\,y\,y'\,a^2}{x}~+~a^2\left[y\,y''~+~(y')^2 \right]} & {~=~} &{0} \\
{~\color{magenta} 7 } &{{\Rightarrow}} &{ \frac{(-1)\,y\,y'}{x}~+~y\,y''~+~(y')^2} & {~=~} &{0} \\
{~\color{magenta} 8 } &{{\Rightarrow}} &{xy\,y''~+~x(y')^2~-~y\,y'} & {~=~} &{0} \\
\end{array}}$
◼ Remarks:
In [(6) magenta color], we use the result from [(4) magenta color] to eliminate $\small{b^2}$
3. We got a differential equation without arbitrary constants.
• We can write:
the differential equation: $\small{xy\,y''~+~x(y')^2~-~y\,y'~=~0}$ represents the given family of curves.
Solved example 25.20
Form the differential equation representing the family of ellipses having foci on the y-axis and center at the origin.
Solution:
1. We know that, the family mentioned in the question is given by the equation: $\small{\frac{x^2}{b^2}~+~\frac{y^2}{a^2}~=~1}$
A few members of the family are shown in the fig.25.13 below:
 |
| Fig.25.13 |
For the green ellipse, a = 3 and b = 5
For the red ellipse, a = 2 and b = 3
For the yellow ellipse, a = 4 and b = 7
2. To eliminate 'a' and 'b', we differentiate the equation as shown below:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\frac{x^2}{b^2}~+~\frac{y^2}{a^2}} & {~=~} &{1} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{\frac{2x}{b^2}~+~\frac{2y\,y'}{a^2}} & {~=~} &{0} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{\frac{x}{b^2}~+~\frac{y\,y'}{a^2}} & {~=~} &{0} \\
{~\color{magenta} 4 } &{{\Rightarrow}} &{x a^2~+~y\,y'\,b^2} & {~=~} &{0} \\
{~\color{magenta} 5 } &{{\Rightarrow}} &{ a^2~+~b^2\left[y\,y''~+~(y')^2 \right]} & {~=~} &{0} \\
{~\color{magenta} 6 } &{{\Rightarrow}} &{ \frac{(-1)\,y\,y'\,b^2}{x}~+~b^2\left[y\,y''~+~(y')^2 \right]} & {~=~} &{0} \\
{~\color{magenta} 7 } &{{\Rightarrow}} &{ \frac{(-1)\,y\,y'}{x}~+~y\,y''~+~(y')^2} & {~=~} &{0} \\
{~\color{magenta} 8 } &{{\Rightarrow}} &{xy\,y''~+~x(y')^2~-~y\,y'} & {~=~} &{0} \\
\end{array}}$
◼ Remarks:
In [(6) magenta color], we use the result from [(4) magenta color] to eliminate $\small{a^2}$
3. We got a differential equation without arbitrary constants.
• We can write:
the differential equation: $\small{xy\,y''~+~x(y')^2~-~y\,y'~=~0}$ represents the given family of curves.
4. This is the same result as in the previous solved example 25.19. The reason can be written in two steps:
(i) The algebraic equation for the two types of ellipses are basically the same. The only difference is that, the arbitrary constants 'a' and 'b' are interchanged.
(ii) In the differential equation, there is no role for the arbitrary constants. So we get the same result.
Solved example 25.21
Form the differential equation representing the family of circles touching the x-axis at the origin.
Solution:
1.
We know that, the circle with center at (a,b) and radius r is given by the
equation: $\small{(x-a)^2~+~(y-b)^2~=~r^2}$
The family mentioned in the question will be similar to the one shown in fig.25.14 below:
 |
| Fig.25.14 |
•
Based on the fig., we can write:
'a' will be zero and 'b' will be equal to 'r'.
•
So the equation of the family becomes:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{x^2~+~(y-r)^2} & {~=~} &{r^2} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{x^2~+~y^2~-~2yr~+~r^2} & {~=~} &{r^2} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{x^2~+~y^2~-~2yr} & {~=~} &{0} \\
\end{array}}$
•
In the above fig.25.14,
♦ r = 1 for the green circle
♦ r = 3 for the red circle
♦ r = 2 for the yellow circle
2. To eliminate 'r', we differentiate the equation as shown below:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{x^2~+~y^2~-~2yr} & {~=~} &{0} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{2x~+~2y\,y'~-~2r\,y'} & {~=~} &{0} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{x~+~y\,y'~-~r\,y'} & {~=~} &{0} \\
{~\color{magenta} 4 } &{{\Rightarrow}} &{x~+~(y~-~r)\,y'} & {~=~} &{0} \\
{~\color{magenta} 5 } &{{\Rightarrow}} &{x~+~\left[y~-~\left(\frac{x^2~+~y^2}{2y} \right)\right]\,y'} & {~=~} &{0} \\
{~\color{magenta} 6 } &{{\Rightarrow}} &{x~+~\left[\frac{2y^2~-~x^2~-~y^2}{2y}\right]\,y'} & {~=~} &{0} \\
{~\color{magenta} 7 } &{{\Rightarrow}} &{x~+~\left[\frac{y^2~-~x^2}{2y}\right]\,y'} & {~=~} &{0} \\
{~\color{magenta} 8 } &{{\Rightarrow}} &{\left[\frac{x^2~-~y^2}{2y}\right]\,y'} & {~=~} &{x} \\
{~\color{magenta} 9 } &{{\Rightarrow}} &{y'} & {~=~} &{\frac{2xy}{x^2~-~y^2}} \\
\end{array}}$
◼ Remarks:
In [(5) magenta color], we use the original equation from [(1) magenta color] to eliminate $\small{r}$
3. We got a differential equation without arbitrary constants.
• We can write:
the differential equation: $\small{y'~=~\frac{2xy}{x^2~-~y^2}}$ represents the given family of curves.
Solved example 25.22
Form the differential equation representing the family of circles touching the y-axis at the origin.
Solution:
1. We know that, the circle with center at (a,b) and radius r is given by the equation: $\small{(x-a)^2~+~(y-b)^2~=~r^2}$
The family mentioned in the question will be similar to the one shown in fig.25.15 below:
 |
| Fig.25.15 |
• Based on the fig., we can write:
'b' will be zero and 'a' will be equal to 'r'.
• So the equation of the family becomes:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{(x-r)^2~+~y^2} & {~=~} &{r^2} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{x^2~-~2xr~+~r^2~+~y^2} & {~=~} &{r^2} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{x^2~+~y^2~-~2xr} & {~=~} &{0} \\
\end{array}}$
• In the above fig.25.15,
♦ r = 1 for the green circle
♦ r = 3 for the red circle
♦ r = 2 for the yellow circle
2. To eliminate 'r', we differentiate the equation as shown below:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{x^2~+~y^2~-~2xr} & {~=~} &{0} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{2x~+~2y\,y'~-~2r} & {~=~} &{0} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{x~+~y\,y'~-~r} & {~=~} &{0} \\
{~\color{magenta} 4 } &{{\Rightarrow}} &{x~+~y\,y'~-~\left(\frac{x^2~+~y^2}{2x} \right)} & {~=~} &{0} \\
{~\color{magenta} 5 } &{{\Rightarrow}} &{\left[x~-~\left(\frac{x^2~+~y^2}{2x} \right) \right]~+~y\,y'} & {~=~} &{0} \\
{~\color{magenta} 6 } &{{\Rightarrow}} &{\left[\frac{2x^2~-~x^2~-~y^2}{2x} \right]~+~y\,y'} & {~=~} &{0} \\
{~\color{magenta} 7 } &{{\Rightarrow}} &{\left[\frac{x^2~-~y^2}{2x} \right]~+~y\,y'} & {~=~} &{0} \\
{~\color{magenta} 8 } &{{\Rightarrow}} &{y\,y'} & {~=~} &{\left[\frac{y^2~-~x^2}{2x} \right]} \\
{~\color{magenta} 9 } &{{\Rightarrow}} &{y'} & {~=~} &{\frac{y^2~-~x^2}{2xy}} \\
\end{array}}$
◼ Remarks:
In [(4) magenta color], we use the original equation from [(1) magenta color] to eliminate $\small{r}$
3. We got a differential equation without arbitrary constants.
• We can write:
the differential equation: $\small{y'~=~\frac{y^2~-~x^2}{2xy}}$ represents the given family of curves.
Solved example 25.23
Form the differential equation representing the family of parabolas having vertex at origin and axis along positive direction of x-axis.
Solution:
1. We know that, the family mentioned in the question is given by the equation: $\small{y^2~=~4ax}$
A few members of the family are shown in the fig.25.16 below:
 |
| Fig.25.16 |
For the green parabola, a = 2
For the red parabola, a = 4
For the yellow parabola, a = 3
2. To eliminate 'a', we differentiate the equation as shown below:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{y^2} & {~=~} &{4ax} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{2y\,y'} & {~=~} &{4a} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{y\,y'} & {~=~} &{2a} \\
{~\color{magenta} 4 } &{{\Rightarrow}} &{y\,y'} & {~=~} &{2\left(\frac{y^2}{4x} \right)} \\
{~\color{magenta} 5 } &{{\Rightarrow}} &{y'} & {~=~} &{\frac{y}{2x}} \\
\end{array}}$
◼ Remarks:
In [(4) magenta color], we use the original equation from [(1) magenta color] to eliminate $\small{a}$
3. We got a differential equation without arbitrary constants.
• We can write:
the differential equation: $\small{y'~=~\frac{y}{2x}}$ represents the given family of curves.
Solved example 25.24
Form the differential equation representing the family of parabolas having vertex at origin and axis along positive direction of y-axis.
Solution:
1. We know that, the family mentioned in the question is given by the equation: $\small{x^2~=~4ay}$
A few members of the family are shown in the fig.25.17 below:
 |
| Fig.25.17 |
For the green parabola, a = 2
For the red parabola, a = 4
For the yellow parabola, a = 3
2. To eliminate 'a', we differentiate the equation as shown below:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{x^2} & {~=~} &{4ay} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{2x} & {~=~} &{4a\,y'} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{x} & {~=~} &{2a\,y'} \\
{~\color{magenta} 4 } &{{\Rightarrow}} &{x} & {~=~} &{y'\left(\frac{x^2}{2y} \right)} \\
{~\color{magenta} 5 } &{{\Rightarrow}} &{1} & {~=~} &{y'\left(\frac{x}{2y} \right)} \\
{~\color{magenta} 5 } &{{\Rightarrow}} &{y'} & {~=~} &{\frac{2y}{x}} \\
\end{array}}$
◼ Remarks:
In [(4) magenta color], we use the original equation from [(1) magenta color] to eliminate $\small{a}$
3. We got a differential equation without arbitrary constants.
• We can write:
the differential equation: $\small{y'~=~\frac{2y}{x}}$ represents the given family of curves.
The link below gives a few more solved examples:
Exercise 9.3
In the next section, we will see solution of differential equations.
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