Question

In this problem you will use variation of parameters to solve the nonhomogeneous equation fy" + 4ty' + 2y = 1 + 12 A. Plug y = p into the associated homogeneous equation (with "0" instead of "13 + 12") to get an equation with only t and n. (Note: Do not cancel out the t, or webwork won't accept your answer!) B. Solve the equation above for n (uset # 0 to cancel out the t). You should get two values for n, which give two fundamental solutions of the form y = t". Y2 = Yi = W(y1, y2) = C. To use variation of parameters, the linear differential equation must be written in standard form y" + py' + qy=g. What is the function g? g(t) = D. Compute the following integrals. Yig dt = W T y28 dt = W E. Write the general solution. (Use c1 and c2 for C1 and C2). y =

Answer 1

salm: - Giver now-homergen ty" +4ty' + 2y on- homogeneous equation at² + t² Let Then y = t. = n thi n-2 y' and y = n(n-1) th Putting y, y, y into the homogeneons part of we get equation with only t and in &n(n-1) + -2 + 4+ (n th-1) + 2th n(n-1) + + 4nth + 2 t In (n-1) + 4x + 2] th oy OY B) As tto n get an equation for [from (2) (n-1) + ant 2 or nn - n tant 2 t zu +2 (nach (+1) OY n=-1, -2 Hence the values for na- 1 and -2 Then we get the fundamental solution as 9,=4" 2 t2 Now 42 t-1 t2 (-2+3 (1) + 2 -- t-4 #[As ta]

Hence in and Ya are linearly independent, a can te written ③ Now equation o 12 y + 2 Y" t + 1 t² componing y" + pyl + 9 g a ZE t , g = til. are P = 1/2 Let the particular integral is y (+) = Alt Y + B (t) Y Alth and B(H) functions of t only. are computed b by formule; Alt) = S- Y = t1 Te = t-2 B(A) = Sto get it at HA) - tt) White) w(4,7.) = -64 Now Alt) and B(A) - 72 glt at W(11991) Therefore ALH - t2 (4+1) dt -t-4 = (+3+ t²) It t4 +3/3 Asi B (t) = St t (t+1) at – ty - ( +4 (3) de - (+ / + +/4) +B, oure A, B, intitrary constants,

th Then the general solution is y=+*++*+($.$)+'+fle = t'tht 2 + + £ - - y = Gt't +6E² 4 - 2 2 t 4 t² 20 where G C are constants. This is the required general solution of the given equation : Ans