Solution of Ex:4.2 Differential Equations with Boundary-Value Problems (7th Edition Book)

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Solution of Ex:4.2 Differential Equations with Boundary-Value Problems (7th Edition Book) consist of Reduction of Order technique. In this post, we will provide solved questions of Ex:4.2 in hard form given below

Quick Overview

Differential equations are used to model many natural phenomena, such as population growth, heat transfer, and fluid flow. In many cases, finding a solution to a differential equation involves finding multiple functions that satisfy the equation. In this article, we will discuss how to use reduction of order to find a second solution to a differential equation.

What is Reduction of Order?

Reduction of order is a technique used to find a second solution to a linear homogeneous differential equation. Furthermore, It is based on the fact that if y1(x) is a known solution to a differential equation of the form:

a(x)y” + b(x)y’ + c(x)y = 0

Then another solution y2(x) can be found by assuming that it has the form:

y2(x) = v(x)y1(x)

Where v(x) is a new function that we need to determine. Furthermore, Substituting this form of y2(x) into the differential equation gives:

a(x)[v”y1 + 2v’y1′ + vy1”] + b(x)[v’y1 + vy1′] + c(x)vy1 = 0

Finally, Simplifying this expression by cancelling out common terms and using the fact that y1(x) satisfies the differential equation, we get:

v”y1 + 2v’y1′ = 0

Finally, this is a second-order linear homogeneous differential equation that can be solved by using the substitution u = v’y1′. We get:

u’ + (y1’/y1)u = 0

This is a first-order linear differential equation that can be solved by separation of variables. Finally, We get:

v(x) = c1∫(exp(-∫(b(x)/a(x))dx)dx) + c2

Where c1 and c2 are arbitrary constants.

Finding the Second Solution to Ex:4.2 Differential Equations with Boundary-Value Problems

Let’s apply the reduction of order technique to find a second solution to the following differential equation:

y” – 4y’ + 4y = 0

Firstly, we can see that

    \[y_1(x) = exp(2x)\]

is a solution of DE. To find the second solution, we assume that it has the form:

y2(x) = v(x)exp(2x)

Secondly, Substituting this form of y2(x) into the differential equation gives:

v”exp(2x) + 4v’exp(2x) = 0

Thirdly, Dividing both sides by exp(2x), we get:

v” + 4v’ = 0

Fourthly, this is a second-order linear homogeneous differential equation that can be solved by using the substitution u = v’. We get:

u’ + 4u = 0

Fifthly, this is a first-order linear differential equation that can be solved by separation of variables. We get:

v(x) = c1exp(-4x) + c2

Sixthly, c1 and c2 are arbitrary constants. Therefore, the second solution to the differential equation is:

y2(x) = (c1exp(-4x) + c2)exp(2x)

Conclusion

Last but not least, Reduction of order is a powerful technique that allows us to find a second solution to a linear homogeneous differential equation. This technique is particularly useful when we already know one solution to the differential equation. By assuming that the second solution has the form of a new function multiplied by the known solution. Finally, we can reduce the problem to a first-order linear differential equation, which can be solved by separation of variables.

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