Question: 4. Solve the following using variation of parameter. (15 points) y ty = exsin (x) step 1 : Homogeneous solution + 9 : + E

 4. Solve the following using "variation of parameter." (15 points) y"ty = exsin (x) step 1 : Homogeneous solution + 9 :+ E r= -0 62- 4(12(17 = 2 ( 1 ] 2r= oti 2 + Bi d = 0 Cost Sinx r= o-iQ - BI B - sinx cosx 41 = cos( X )

yz = sin(x ) w (y , , 4 2 ) =17 0 Uh = CI cos ( x) + cz sin (x) Step 2: particular solution using "variation of parameter" Up = uisin(x ) + uzcos (x) yp' = ui sin ( x )+ UIcos ( x ) + uz cos ( x) - uzSin

4. Solve the following using "variation of parameter." (15 points) y" ty = exsin (x) step 1 : Homogeneous solution + 9 : + E r= -0 62- 4(12(17 = 2 ( 1 ] 2 r= oti 2 + Bi d = 0 Cost Sinx r= o-i Q - BI B - sinx cosx 41 = cos( X ) yz = sin(x ) w (y , , 4 2 ) = 17 0 Uh = CI cos ( x) + cz sin (x ) Step 2: particular solution using "variation of parameter" Up = uisin (x ) + uzcos (x) yp' = ui sin ( x ) + UIcos ( x ) + uz cos ( x) - uzSin (x ) Assume : hisin ( x ) + uz cos ( x ) = 0 Up = UIcos ( x) - uzsin(x) Up " = ui cosix ) - uisin (x ) - uz sincx) - uzcos (x)4" + y = ex sincx ) hi cos (x ) - usin (x ) - uz sin (x ) - 12tos(x ) + uisintx ] + uzcostx ) = ex sincx ) UI cosx - uz sin ( x) = ex sincx) hisin ( x ) + uz cos(x ) = 0 UI cosx - Uzsin (x) = ex sin(x) hisin( x ) + us fin (x ) . cos(x ) = 0 UI cos(x ) - ux fin(x ) . cos(x ) = ex sin ( x ) cos ( x ) UI ' [ sing ? ( x ) + Cos = ( x ) ] = ex sin ( x ) cos(x ) U = ex sin ( x ) cos ( x) u = ex sin ( x ) cos ( x ) d x = 2 ex sin (zx ) dx Computer = 5 er cos (2 x ) + 10 ex sin (2x )UI SHATX ) LOS( X ) + 42 cos ( x ) = - u , sintx ) Cos ( x ) + 4 2 sin ( x ) = - ex sin ( x ) U2 [ cos ( x ) + sin "(x ) J = - ex sin?(x ) U2 = - ex sin? (x ) U2 = - ex sin ? ( x ) dx Computer = -exsin?(x) - - ex s (2x) + ex sin (2x) yp = uIsin ( x ) + uzcos ( x ) yP - 5 e * cos 12 x ) + 10 e " sin (2 x ) sin ( x ) + - exsin ? ( x ) - els ( 2 x ) + ex sin ( 2x ) . cos( * ) - $ ex (2605() - 1 ) + (25in(x) cos(x ) . sin (x ) + - exsin ? (x ) - ex ( 1 - 2 sin x ) ) . cos ( x ) 1 5 ex 2 sin ( * ) cos ( x ) . (5 ( x )= 2 ex sintx ) los ( x ) + 5 ex sin ( x ) 5 ex sintx , Cos( x ) - ex sinful Cos( x ) 5 ex cos( x ) + ex sintx ) cos( x ) + 2 ex sintx ) cos (x ) 5 ex sin ( x ) - ex cs ( x ) step 3 : General solution : y = yu + up = CI COS ( X ) + C z Sin ( x) + ex sin (x) - - ex cos (x)f0 Solve the following differential equation (nd the general solution) using the method of \"Variation of Parameter". Show all your work. You must use variation of parameter if you use any other method. you will get no credit. (10 points) '//_2'/ ,2 ' y y+y 1+t'3 You cannot use any formula for the variation of parameters. You must do the way we did in the class. Make sure to review and rewatch the recorded video on this topic before proceeding. If you use any formula, you will get 0

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