Higher-Order Non-Homogeneous DEs - Examples & Applications

Conceptual Examples: Superposition Principle

Superposition Principle (Conceptual)

Suppose yp1=3e2xy_{p1} = 3e^{2x} is a particular solution to y′′+4y=12e2xy'' + 4y = 12e^{2x} and yp2=x2−1y_{p2} = x^2 - 1 is a particular solution to y′′+4y=4x2−2y'' + 4y = 4x^2 - 2. Find a particular solution to the equation y′′+4y=6e2x−4x2+2y'' + 4y = 6e^{2x} - 4x^2 + 2.

Superposition Conceptual Solution

0 of 3 Steps Completed
1

Method of Undetermined Coefficients

Undetermined Coefficients Problem 1 (Standard Form)

Find the general solution of:
y′′−3y′−4y=3e2xy'' - 3y' - 4y = 3e^{2x}

Undetermined Coefficients Solution 1

0 of 4 Steps Completed
1

Undetermined Coefficients Problem 2 (Multiplication Rule)

Find the general solution of:
y′′−2y′+y=exy'' - 2y' + y = e^x

Undetermined Coefficients Solution 2

0 of 4 Steps Completed
1

Undetermined Coefficients (Trig Function)

Find the general solution of:
y′′+4y=3sin⁡(x)y'' + 4y = 3\sin(x)

Trig Undetermined Solution

0 of 4 Steps Completed
1

Annihilator Approach

Annihilator Approach Problem 1

Determine the annihilator operator for the function f(x)=x2e3xf(x) = x^2 e^{3x}.

Annihilator Operator Solution

0 of 3 Steps Completed
1

Annihilator Approach Problem 2

Solve y′′−y′=4xy'' - y' = 4x using the Annihilator Approach.

Annihilator Solution 2

0 of 5 Steps Completed
1

Variation of Parameters

Variation of Parameters Problem 1

Find the particular solution of:
y′′+y=sec⁡xy'' + y = \sec x

Variation of Parameters Solution 1

0 of 4 Steps Completed
1

Variation of Parameters Problem 2

Find the general solution of:
y′′−2y′+y=exx2+1y'' - 2y' + y = \frac{e^x}{x^2 + 1}

Variation of Parameters Solution 2

0 of 4 Steps Completed
1
PreviousHigher-Order Non-Homogeneous DEs - Theory & Concepts
Quiz Me
NextApplications of Higher-Order DEs - Theory & Concepts