Differential equations practice problems with solutions pdf

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General and Particular Solution of Differential Equation

The solution of a differential equation is the relationship between the variables included which satisfies the differential equation. There are two types of solutions of differential equations namely, the general solution of differential equations and the particular solution of the differential equations. The general and the particular solutions of differential equations make use of some steps of integration to solve the equations. 

What is the Differential Equation?

A differential equation is an equation that includes one or more terms and also includes the derivatives of one variable (i.e., dependent variable) in terms of the other variable (i.e., independent variable)dt/dz = f(z). Here “z” is an independent variable and “t” is a dependent variable. For example, dt/dz = 5z.

Methods to Solve Differential Equations

There are 5 methods to solve the differential equation. These 5 Methods Are:

1. Solution by inspection

2. Variable separable

3. Homogeneous

4. Linear differential equation

5. General

General Solution of  a Differential Equation

A General Solution of the nth order differential equation is defined as the solution that includes an important arbitrary constants. It is necessary for us to introduce an arbitrary constant as soon as integration is performed if we solve a first-order differential equation by a variable method. Hence, you can see after simplifying that the solution of the differential equation of first order includes an important arbitrary constant.

Similarly, the general solution of a second-order differential equation will include important arbitrary constants and so on. Geometrically, the general solution represents an n-parameter family of curves. For example, the general solution of the differential equation dy/dx = 8x² which is found to be y = x³ + C,  where c is considered as  an arbitrary constant, represents a one-parameter family of curves as shown in the figure given below.

Particular Solution of  a Differential Equation

The particular solution of a differential equation is a solution which we get from the general solution by giving particular values to an arbitrary solution. The conditions for computing the values of arbitrary constants can be given to us in the form of an initial-value problem or Boundary Conditions depending on the questions.

Singular Solution of a Differential Equation

The singular solution of a differential equation is a special kind of particular solution of a differential equation but it cannot be derived from the general solution of a differential equation by assigning the values of the random constant.

Terms related to Differential Equations

• Order: Differential equation which has been differentiated for the highest number of times.

• Degree: the power of a variable of the highest order derivative when the differential equation is in the polynomial form.

• Partial Differential Equation: a differential equation with a function of several variables of their partial derivatives is called a partial differential equation.

Note: Order and degree of a differential equation are always positive integers.

How to form a Differential Equation from a given equation?

Let x and y be the independent variable and the dependent variable respectively for an equation in which “k” is an arbitrary constant. To form a differential equation from the given equation, follow the steps:

• Check the number of arbitrary constants,

• Differentiate the given equation equal to the number of arbitrary constants, for example, if the number of arbitrary constants in an equation is “m”, then differentiate the equation for “m” number of times.

The equation we obtain by the subsequent differentiation is our required differential equation.

Methods of solution

The five methods of solution are:

1. Solution by Inspection: If the differential equation is of the form f(f1(x,y))d(f1(x,y))+g(f2(x,y))d(f2(x,y))+……=0, then each term can be separately integrated. 

2. Variable Separation Method: If a differential equation can be written in a form such that variables are separated for integrating then the equation can be solved. This type of equation can be written as y’ = f(x)g(y) where y’ is the differentiation of x with respect to y.

3. Homogeneous Differential Equation: An equation of the form y’= dy/dx = f(x,y)/g(x,y) where f and g are homogeneous then such a differential equation is known as homogeneous differential equation. 

4. Linear Differential Equation: A differential equation of the form dy/dx + Py = Q where P and Q are arbitrary constants or functions of x, is known as a linear differential equation. 

5. General Solution: A general solution of a differential equation is an equation with dependent variables in terms of independent variables. The general equation can be given as 

Differential Equation Practice Problems

Here, you can see some of the differential equation practice problems.

1. Find the general solution of the following differential equation dt/dx = (1 + x²) ( 1+ t²)

Solution: The given differential equation is dt/dx = (1 + x²) ( 1+ t²)

dt/( 1+ t²) = (1 + x²)/dx

By integrating both sides of the above equation, we get

∫dt/( 1+ t²) = ∫(1 + x²)/dx

tan-1t = ∫dx +∫dx²

tan-1t  = x + x³/3 + C

The above equation is the required general solution of the differential equation.

1. Find the general solution of the differential equation given below \[ \frac {dt}{dx} = e^{z + t}\]

Solution: We have, \[ \frac {dt}{dx} = e^{z + t}\]

Using the law of exponent, we get dt/dz = \[e^{z} + e^{t}\]

By separating variables by variable separable procedure, we get \[e^{-t} dt =  e^{z} dz\] 

Now taking integration of both the side, we get \[e^{-t} dt =  e^{z} dz\]

On integrating, we get

\[- e^{-t} = e^{z} + C\]

\[e^{z} + e^{-t} = -C or e^{z} + e^{-t} = C\]

Differential Equation Practice Problems With Solutions

Here, you can see some of the differential equation practice problems with solutions.

1. Find the particular solution of a differential equation which satisfies the below condition \[\frac{dy}{dx} = \frac{1}{x^{2}}; y(1) = 4\]

Solution: We will first find the general solution of a differential equation. To do this, we will integrate both sides to find y 

\[\frac{dy}{dx} = \frac{1}{x^{2}};\]

\[ y(1) = 4\]

\[ y = \frac{-1}{x + c}\]

\[ y = \int (\frac{1}{x^{2}}) dx\]

\[ y = \int (\frac{1}{x^{2}}) dx\]

\[ y = \frac{x^{-1}}{-1 + c}\]

Now, we apply our initial conditions (x = 1, y = 4) and solve for C, which will give us our particular solution:

\[ 4 = {-1 + c}\]

Now, we will solve for

\[ 4 = {-1 + c}\]

5 = C

\[ y = \frac{-1}{x + 5}\]

Hence, the particular solution of a differential equation is y = -1/x + 5