- The Given Differential Equation Is Already In Standard Form As The Coefficient Of The Highest Order Derivative Is 1 Y 1 (19.01 KiB) Viewed 33 times
The given differential equation is already in standard form, as the coefficient of the highest order derivative is 1. y′′+3y′+2y=2+ex1 The next Wronskians to calculate use y1=e−x and y2=e−2x that we identified from the complementary function and the function of x that makes the equation nonhomogeneous, f(x)=2+ex1. W1=∣∣0f(x)y2y2′∣∣=∣∣0(2+ex)−1e−2x−2e−2x∣∣ = W2=∣∣y1y1′0f(x)∣∣ =∣∣e−x−e−x0(2+ex)−1∣∣ =
Solve the differential equation by variation of parameters. y′′+3y′+2y=2+ex1 Step 1 We are given a nonhomogeneous second-order differential equation. Similar to the method of solving by undetermined coefficients, we first find the complementary function yc for the associated homogeneous equation. This time, the particular solution γp is based on Wronskian determinants and the general solution is y=yc+yp. First, we must find the roots of the auxiliary equation for y′′+3y′+2y=0. m2+3m+2=0 Solving for m, the roots of the auxiliary equation are as follows. smaller value m1=−2 larger value m2= Step 2 We have found that the roots of the auxiliary equation are m1=−1 and m2=−2. In the case that the auxiliary equation for a second-order linear equation has two distinct, real roots, we know the complementary function is of the form yc=c1em1x+c2em2x. Therefore, the complementary function is as follows. yc=c1e−x+c2e−2x Let y1=e−x and y2=e−2x be the two independent solutions which are terms of the complementary function. We will find functions u1(x) and u2(x) such that yp=u1y1+u2y2 is a particular solution. These new functions are found by calculating multiple Wronskians. First, find the following Wronskian. w(y1(x),y2(x))=W(e−x,e−2x) ∣∣y1(x)y1′(x)y2(x)y2′(x)∣∣=∣e−xe−2x