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The differential equation of a family of curves(this curve can be parabola, ellipse, circles, etc.) is differentiated as many times as the number of arbitrary constants in the equation. This means that the order of the differential equation we get from any family of curves is the number of arbitrary constants in the equation of the curve given.
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Key Takeaways: Differentiation, Equation, Curve, Differential, Solution, Ellipse, Parabola, Circle, Differential equation
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Differential Equation Solution of Differential Equations
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A differential equation is a mathematical equation that connects one or more functions and their derivatives. In applications, functions are used to represent physical quantities, derivatives are used to describe their rates of change, and the differential equation is used to define a relationship between them.
Differential equations are commonly utilized in biology, physics, engineering, and other sciences. The differential equation's major goal is to investigate the solutions that satisfy the equations as well as the properties of the solutions.
The highest order of the derivative appearing in a differential equation is the order of the equation.
For example: The equation is given to you is x2 + y2 = r2 which is the equation of concentric circles with the centre as origin so differentiating the given equation with respect to x, we get:
2x + 2y dy/dx = 0
dy/dx = -x/y
dy/dx +x/y = 0 is the equation of all concentric circles with its center at origin. The number of arbitrary constants in our given equation x2 + y2 = r2 is 1 which is r. Therefore, differentiating the equation once is enough.
Also Read: First Order Differential Equation
Solution of Differential Equations
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The solution of differential equations is of two types-general and particular. The general solution becomes the specific solution of the problem when the arbitrary constant of the general solution obtains a unique value.
The particular solution of a differential equation is obtained by employing the boundary conditions (also known as the initial conditions).
A Particular Solution is a differential equation solution that is derived from the General Solution by assigning particular values to the arbitrary constants. Depending on the query, the requirements for finding the values of the random constants can be submitted to us as an Initial-Value Problem or Boundary Conditions.
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Solved Examples
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In order to properly understand the concept, we will try to find differential equations when the general solution is given so we will look at two examples.
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Example 1
Problem Statement: Form differential equation representing the family of ellipses whose center is at origin and focus on the x-axis.
Solution: The equation representing the ellipse whose center is at origin and focus on the x-axis can be represented as
x2 / a2 + y2 / b2 = 1 where a > b as focus is on x-axis.
The above equation represents an ellipse whose center is at origin and focus on the x-axis but we need the equation of the family of ellipses which can cover all the ellipses possible.
The equation has 2 arbitrary constants and therefore we must differentiate the equation with respect to x twice to get rid of the arbitrary constants and get our differential equation,
Also Read: Integration by Partial Function
First differentiation
2x / a2 + dy/dx * 2y / b2 = 0
dy/dx * 2y / b2 = - 2x / a2
Canceling 2 from both sides,
dy/dx * y / x = -b2 / a2
Second differentiation
dy/dx * y / x = -b2 / a2
(y/x)’ dy/dx + (y/x) d2y/dx2 =0 is the required differential equation
Also Read: Limit Formula
Example 2
Problem Statement: Form differential equation representing the family of parabolas whose vertex is at origin and axis along the positive direction of the x-axis.
Solution: The equation representing the parabolas whose vertex is at origin and axis along the positive direction of the x-axis
y2 = 4ax
The above equation represents a parabola whose vertex is at origin and axis along the positive direction of the x-axis but we need the equation of the family of parabolas which can cover all the parabolas possible.
The equation has only 1 arbitrary constant and therefore we must differentiate the equation with respect to x only once to get rid of the arbitrary constant and get our differential equation,
First differentiation
y2 = 4ax
y2 / x = 4a
(2y * dy/dx - y2)/x2 = 0 is the required differential equation
Also Read:
Things to Remember
- When finding out the order of the differential equation, one must try to notice the highest order of Derivatives used in the complete equation.
- In order to construct the differential equation when a general solution is given, we have to carefully notice the number of arbitrary constants used in the equation and finding out the number of arbitrary constants is of utmost importance.
- While preparing this topic of differential equations, one must be absolutely well versed with all the equations as well as the geometry of the curves like a parabola, hyperbola, circles, ellipse, etc.
- The earlier steps of these questions are most crucial as if you calculate the number of arbitrary constants wrong and differentiate the equation more/less than actually necessary then your whole question becomes wrong.
- When differentiating, do not blindly start differentiating the question but always set up the equation by moving elements of the equation from the left-hand side of the equal sign to the right-hand side or vice versa. This setting up is absolutely essential to save time and make the calculations easier.
Also Read: Tan2x Formula
Sample Questions
Ques. Form differential equations representing the family of curves by eliminating the arbitrary constants ‘a’ and ‘b’ → x/a + y/b =1. (2 marks)
Ans. The equation representing the curve x/a + y/b =1 has two arbitrary constants and therefore we must differentiate the given equation twice to get the required differential equation.
Differentiating both sides with respect to x,
First differentiation
1 / a + 1/b (dy/dx) = 0
Again differentiating both sides with respect to x,
Second differentiation
0 + 1/b (d2y/dx2) = 0
d2y/dx2 = 0 is the required differential equation
Ques. Form differential equations representing the family of curves by eliminating the arbitrary constants ‘a’ and ‘b’ → y2 = a(b2 - x2). (3 marks)
Ans. The equation representing the curve y2 = a(b2 - x2) has two arbitrary constants and therefore we must differentiate the given equation twice to get the required differential equation.
First, simplifying the equation by multiplying arbitrary constant on right hand side,
y2 = ab2 - ax2
Differentiating both sides with respect to x,
First differentiation
2y(dy/dx)= 0 - a(2x)
Canceling 2 from both sides.
y(dy/dx) = -ax
Using dy/dx as y`
yy`=-ax…………………….(1)
Again differentiating both sides with respect to x,
Second differentiation
Using d2y/dx2 as y``
yy``+ y`y` = -a…………………….(2)
We have to substitute the value of -a from equation (2) in equation (1)
Putting (2) in (1),
yy`=(yy``+(y`)2).x
yy` = xyy`` + x(y`)2
xyy`` + x(y`)2 - yy` is the required differential equation
Ques. Form differential equations representing the family of curves by eliminating the arbitrary constants ‘a’ and ‘b’ → y = ae3x + be-2x). (5 marks)
Ans. The equation representing the curve y = ae3x + be-2x has two arbitrary constants and therefore we must differentiate the given equation twice to get the required differential equation.
y = ae3x + be-2x…………………….(1)
Using dy/dx as y`
Differentiating both sides with respect to x,
First differentiation
y` = 3ae3x -2be-2x…………………….(2)
Multiplying (1) with 3, we get
3y = 3ae3x + 3be-2x…………………….(3)
Subtracting (3) from (2),
y` - 3y = 3ae3x -2be-2x - ae3x - 3be-2x
y` - 3y = -5be-2x…………………….(4)
Again differentiating both sides with respect to x,
Second differentiation
Using d2y/dx2 as y``
y`` - 3y ` = 10be-2x…………………….(5)
Multiplying (4) with 2, we get
2y` - 6y = -10be-2x…………………….(6)
Adding (5) and (6)
y`` - 3y ` + 2y` - 6y= 0
y``-y`-6y=0
xy``-y`-6y=0 is the required differential equation
Ques. Form differential equations representing the family of curves by eliminating the arbitrary constants ‘a’ and ‘b’ → y = e2x (a + bx). (5 marks)
Ans. The equation representing the curve y = e2x (a + bx) has two arbitrary constants and therefore we must differentiate the given equation twice to get the required differential equation.
y =e2x (a + bx)…………………….(1)
Using dy/dx as y`
Differentiating both sides with respect to x,
First differentiation
y` = e2x (0 + b(1)) + 2e2x (a + bx)
y` = be2x + 2e2x (a + bx)
y` = be2x + 2y ( From (1) ) …………………….(2)
Again differentiating both sides with respect to x,
Second differentiation
Using d2y/dx2 as y``
y`` = 2be2x + 2y`…………………….(3)
Multiplying (2) with 2, we get
2y` = 2be2x + 4y…………………….(4)
Subtracting (4) from (3)
y`` - 2y` = 2be2x + 2y` -2be2x - 4y
y`` - 2y`=2y`- 4y
y`` - 4y` + 4y=0 is the required differential equation
Ques. Form differential equations representing the family of curves by eliminating the arbitrary constants ‘a’ and ‘b’ → y = ex (acosx + bsinx). (5 marks)
Ans. The equation representing the curve y = ex (acosx + bsinx) has two arbitrary constants and therefore we must differentiate the given equation twice to get the required differential equation.
y = ex (acosx + bsinx)…………………….(1)
Using dy/dx as y`
Differentiating both sides with respect to x,
First differentiation
y` = ex (-asinx + bcosx) + ex (acosx + bsinx) [From (1)]
y` = ex (-asinx + bcosx) +y…………………….(2)
Again differentiating both sides with respect to x,
Second differentiation
Using d2y/dx2 as y``
y`` = ex (-acosx -bsinx) +ex (-asinx + bcosx)+y` [From (2)]
As (2) can be represented as
y` - y= ex (-asinx + bcosx)…………………….(3)
y`` = -ex (acosx +bsinx) + (y` - y) + y` [From (3)]
ex (acosx +bsinx) is y, therefore,
y`` = -y + (y` - y) + y`[From (1)]
y`` = 2y` -2y
y`` - 2y` + 2y=0 is the required differential equation
Ques. Form differential equations representing the family of circles touching the y-axis at origin. (3 marks)
Ans. The general equation of a circle is (x-a)2 + (y-b)2 = r2 where the center is at (a,b) and radius r.
Let the equation of circle having center (a,0) and radius a is:
(x-a)2 + (y-0)2 = a2
x2 + a2 -2ax + y2 = a2
x2 + y2 = 2ax…………………….(1)
The equation given has one arbitrary constant and therefore we must differentiate the given equation only once to get the required differential equation.
First differentiation
Differentiating both sides with respect to x,
Using dy/dx as y`
2x + 2yy` = 2a…………………….(2)
We have to substitute the value of 2a from equation (2) in equation (1)
Putting (2) in (1),
x2 + y2=(2x + 2yy`)x
x2 + y2 = 2x2 + 2xyy`
2xyy` + x2 = y2 is the required differential equation
Ques. Form differential equation representing the family of parabolas whose vertex is at origin and axis along the positive direction of the y-axis. (4 marks)
Ans. The equation representing the parabolas whose vertex is at origin and axis along the positive direction of the x-axis
x2 = 4ay…………………….(1)
The above equation represents a parabola whose vertex is at origin and axis along the positive direction of the y-axis but we need the equation of the family of parabolas which can cover all the parabolas possible.
The equation has only 1 arbitrary constant and therefore we must differentiate the equation with respect to x only once to get rid of the arbitrary constant and get our differential equation,
First differentiation
x2 = 4ay
2x = 4a * dy/dx
Using dy/dx as y`
4a=2x/y`…………………….(2)
Putting (2) in (1),
x2 =2x/y` X y
xy`=2y
xy`-2y = 0 is the required differential equation
Ques. Form differential equation representing the family of ellipses whose center is at origin and foci on the y-axis. (4 marks)
Ans. The equation representing the ellipse whose center is at origin and focus on the y-axis can be represented as
x2 / b2 + y2 / a2 = 1…………………….(1)
where a > b as focus is on the y-axis.
The above equation represents an ellipse whose center is at origin and focus on the y-axis but we need the equation of the family of ellipses which can cover all the ellipses possible.
The equation has 2 arbitrary constants and therefore we must differentiate the equation with respect to x twice to get rid of the arbitrary constants and get our differential equation,
First differentiation
2x / b2 + dy/dx X 2y / a2 = 0
Using dy/dx as y`
2(x/b2 +yy`/a2)=0
x/b2 +yy`/a2=0…………………….(2)
Multiply (2) by x,
x2/b2 +xyy`/a2=0…………………….(3)
Subtract (1) from (3)
xyy`/a2 - y2 / a2 = -1
(xyy`/a2 - y2)/a2 =-1
xyy`/a2 - y2 =-a2
Second differentiation
xy.y``+xy`.y`+y.y`-2yy`=0
xy.y``+xy`.y`+y.y`-2yy`=0
xy.y``+x(y`)2+-yy`=0 is the required differential equation.
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