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Formation of Differential Equation and Solutions of a Differential Equation - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

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  • Formation of Differential Equation and Solutions of a Differential Equation is considered one of the most asked concept.

  • 26 Questions around this concept.

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The differential equation for the family of curves x^{2}+y^{2}-2ay=0, where a is an arbitrary constant is.

The differential equation which represents the family of curves $y= c_{1}e^{c_{2}x},$ where $c_{1}\: and\: c_{2}$ are arbitrary constants, is

If the differential equation representing the family of all circles touching $x$-axis at the origin is $\left(x^2-y^2\right) \frac{d y}{d x}=g(x) y$, then $g(x)$ equals :

The differential equation whose solution is Ax^{2}+By^{2}=1,\; where\; A\; and\; B are arbitrary constant, is of

The differential equation of all circles passing through the origin and having their centres on the x-axis is.

The differential equation of the family of circles with fixed radius 5 units and centre on the line y=2 is

The degree and order of the differential equation of the family of all parabolas whose axis is x-axis, are respectively.

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The solution of the equation  \frac{d^{2}y}{dx^{2}}=e^{-2x}.

Concepts Covered - 1

Formation of Differential Equation and Solutions of a Differential Equation

Formation of Differential Equation

We have studied the general formula of the parabola which is $y^2=4 a x$

This equation represents a family of parabolas with an arbitrary constant. With different values, we get a different parabola in this family.

To form its D.E, let us first differentiate it w.r.t. x,

$
2 \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}=4 \mathrm{a} \quad \text { or } \quad \frac{\mathrm{y}}{2} \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{a}
$

Putting this value of 'a' in the original equation of a parabola
$
\begin{aligned}
& y^2=4 \cdot \frac{y}{2} \frac{d y}{d x} \cdot x \\
& 2 x \frac{d y}{d x}=y
\end{aligned}
$

This is the Differential Equation for a family of parabolas $y^2=4 a x$

Note that there is one arbitrary constant in the original equation, and the order of the D.E. formed is also 1. This is an important result: if an equation contains n arbitrary constants, then its D.E. will have an order equal to n.

If we are given a relation between the variables x, y with n arbitrary constants C1, C2, .... Cn, then to form its D.E., we differentiate the given relation n times in succession with respect to x, we have n + 1 equations altogether. Now using these we eliminate the n arbitrary constants. The result is a differential equation of the nth order. 

Consider the equation of the family of ellipses with a and b as arbitrary constants
$
\frac{x^2}{a^2}+\frac{y^2}{b^2}=1
$

Differentiate Eq (4) w.r.t. ' $x$ '
$
\frac{2 \mathrm{x}}{\mathrm{a}^2}+\frac{2 \mathrm{y}}{\mathrm{~b}^2} \cdot\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)=0 \quad \text { or } \quad-\frac{\mathrm{b}^2}{\mathrm{a}^2}=\frac{\mathrm{yy}^{\prime}}{\mathrm{x}}
$

Differentiate above equation w.r.t. 'x'
$
\begin{gathered}
0=\frac{\mathrm{x}\left(\mathrm{y}^{\prime 2}+\mathrm{yy}^{\prime \prime}\right)-\mathrm{yy}^{\prime}}{\mathrm{x}^2} \\
\mathrm{yy}^{\prime}=\mathrm{xyy}^{\prime \prime}+\mathrm{x}\left(\mathrm{y}^{\prime}\right)^2
\end{gathered}
$

This is the D.E. that represents the given family of ellipses

Note: If arbitrary constants appear in addition, subtraction, multiplication, or division, then we can club them to reduce into one new arbitrary constant. Hence, the differential equation corresponding to a family of curves will have the order exactly same as a number of essential arbitrary constants (number of arbitrary constants in the modified form) in the equation of the curve.

Illustration : 

Differential equation of the equation $\mathrm{y}=(\mathrm{a}+\mathrm{b}) \mathrm{e}^{\mathrm{x}}+\mathrm{e}^{\mathrm{x}+\mathrm{c}}$ is Here, the number of the arbitrary constant is $3: a, b, a n d$. But we can club arbitrary constants together
$
\mathrm{y}=\left(\mathrm{a}+\mathrm{b}+\mathrm{e}^{\mathrm{c}}\right) \mathrm{e}^{\mathrm{x}}
$
which is of the form, $\mathrm{y}=\mathrm{Ae}^{\mathrm{x}} \quad$ where, $\mathrm{A}=\left(\mathrm{a}+\mathrm{b}+\mathrm{e}^{\mathrm{c}}\right)$ hence, corresponding Differential equation will be of order 1
$
\begin{aligned}
& \Rightarrow \frac{d y}{d x}=A e^x \\
& \Rightarrow \frac{d y}{d x}=y
\end{aligned}
$

Solution of D.E.

If the given D.E. is 

$
\begin{aligned}
& 2 x \frac{d y}{d x}=y \\
& 2 \frac{d y}{y}=\frac{d x}{x} \\
& 2 \int \frac{d y}{y}=\int \frac{d x}{x} \\
& 2 \ln (y)=\ln (x)+c
\end{aligned}
$

As c is a constant of integration, so it can take any real value. We can write it as $\ln (4 \mathrm{a})$ as well, where a is now the arbitrary constant.
$
\begin{aligned}
& \ln \left(y^2\right)=\ln (x)+\ln (4 a) \\
& y^2=4 a x
\end{aligned}
$

This is the general solution of the differential equation, which represents the family of the parabola

If we put some real value of an in this equation, then we get a particular solution of the given differential equation. For example,$\mathrm{a}=2$, then $\mathrm{y}^2=8 \mathrm{x}$ is a particular solution of this equation.

Hence, the solution of the differential equation is a relation between the variables of the equation that satisfy the D.E. and does not contain any derivatives or any arbitrary constants.

A general solution of a differential equation is a relation between the variables (not involving the derivatives) which contains the same number of the arbitrary constants as the order of the differential equation. 

A particular solution of the differential equation is obtained from the general solution by assigning particular values to the arbitrary constant in the general solution.

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Formation of Differential Equation and Solutions of a Differential Equation

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Formation of Differential Equation and Solutions of a Differential Equation

Mathematics for Joint Entrance Examination JEE (Advanced) : Calculus

Page No. : 10.2

Line : 23

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