Define 2nd order linear homogenous/non-homogenous differential equations along with two examples?

Name: Define 2nd order linear homogenous/non-homogenous differential equations along with two examples?
Uploaded: 2022-02-23T17:22:57+00:00
Description: Define 2nd order linear homogenous/non-homogenous differential equations along with two examples?

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Lakisha Archer · Accepted Answer

Step 1Consider the second order differential equation of the form y″+p(x)y′+q(x)y=f(x) where p(x),q(x),f(x)  are continuous functions .The equation is known as non homogeneous differential equation.If f(x)=0, the equation becomes y″+p(x)y′+q(x)y=0Homogeneous linear differential equations is of the type y″+p(x)y′+q(x)y=0 where p(x),q(x)  are continuous functions .The characteristic equation can be written as t2+pt+q=0The general solution of the equation depends upon the roots of the characteristic equation.1. If the roots are real and distinct then the solution of the differential equation is y=c1ek1x+c2ek2x where k1 and k2 are the roots of the characteristic equation. c1 and c2 are arbitrary constants.2. If the roots are real and distinct then the solution of the differential equation is y(c1x+c2ek1)x where k1 is the  repeated  root of the characteristic equation. c1 and c2 are arbitrary constants.3. If the roots are complex then the solution of the differential equation is y=eαx(c1cos⁡βx+c2sin⁡βx) where α+βi and α−βi are the complex root of the characteristic equation. c1 and c2 are arbitrary constants.Solution of Non homogenous differential equation is calculated as follows.The solution of Non homogenous differential equation has two parts.The solution is given by y=yc+ypwhere yc is calculated the same way as in homogenous equation. yp is the particular solution.Step 2example 1:consider the differential equation y″+36y=0the characteristic equation is t2+25=0solving the equationt2=−36t=±6itherefore, y=0±5ithe roots are complex Therefore, the solution is y=e0x(c1cos⁡5x+c2sin⁡5x)Simplify to get the solution y=(c1cos⁡5x+c2sin⁡5x) where c1 and c2 are arbitrary constants.Step 3Example 2:consider the differential equation y2y′−3y=e2tfind the general solution for the equation y″−2y′−3y=0the characteristic equation is t2−2t−3=0t2−2t−3=0

Thomas White · Accepted Answer

Solution.We will use the method of undetermined coefficients. The right side of the given equation is a linear function f(x)=ax+b. Therefore, we will look for a particular solution in the formy1=Ax+BThen the derivatives arey′1=A,y=0.Substituting this in the differential equation gives:0+A−6(Ax+B)=36x,⇒A−6Ax−6B=36x.The last equation must be valid for all values of  so the coefficients with the like powers of  in the right and left sides must be identical:−6A=36A−6B=0We find from this system that A=−6,B=−1. As a result, the particular solution is written asy1=−6x−1.Now we find the general solution of the homogeneous differential equation. Calculate the roots of the auxiliary characteristic equation:k2+k−6=0,⇒D=1−4⋅(−6)=25,⇒k1,2=−1±252=−1±52=−3,2.Hence, the general solution of the related homogeneous equation is given byy0(x)=C1e−3x+C2e2xThus, the general solution of the initial nonhomogeneous equation is expressed by the formulay=y0+y1=C1e−3x+C2e2x−6x−1.

Vasquez · Accepted Answer

Consider the second order differential equation of the form y&#039;&#039;+p(x)y&#039;+q(x)y=f(x) where p(x),q(x),f(x) are continuous functions .The equation is known as non homogeneous differential equation.If f(x)=0, the equation becomes y&#039;&#039;+p(x)y&#039;+q(x)y=0Homogeneous linear differential equations is of the type y&#039;&#039;+p(x)y&#039;+q(x)y=0 where p(x),q(x) are continuous functions .The characteristic equation can be written as t2+pt+q=0The general solution of the equation depends upon the roots of the characteristic equation.1. If the roots are real and distinct then the solution of the differential equation is y=c1ek1x+c2ek2x where k1 and k2 are the roots of the characteristic equation. are arbitrary constants.2. If the roots are real and distinct then the solution of the differential equation is y(c1x+c2ek1)x where k1 is the repeated root of the characteristic equation. c1 and c2 are arbitrary constants.3. If the roots are complex then the solution of the differential equation is y=eαx(c1cos⁡βx+c2sin⁡βx) where α+βi and α−βi are the complex root of the characteristic equation. c1 and c2 are arbitrary constants.Solution of Non homogenous differential equation is calculated as follows.The solution of Non homogenous differential equation has two parts.The solution is given by y=yc+ypwhere yc is calculated the same way as in homogenous equation.yp is the particular solution.consider the differential equation y″+36y=0the characteristic equation is t2+25=0solving the equationt2=−36t=±6itherefore, y=0±5ithe roots are complexTherefore, the solution is y=e0x(c1cos⁡5x+c2sin⁡5x)Simplify to get the solution y=(c1cos⁡5x+c2sin⁡5x) where c1 and c2 are arbitrary constants.consider the differential equation y&quot;−2y′−3y=e2tfind the general solution for the equation y″−2y′−3y=0the characteristic equation is t2−2t−3=0t2−2t−3=0t2−3t+t−3=0t(t−3)+1(t−3)=0(t+1)(t−3)=0The solution of the characteristic equation are t=-1,t=3S which are real and distinct.Therefore, the solution the differential equation is y=c1e−1t+c2e3t

Define 2nd order linear homogenous/non-homogenous differential equations along with two

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