Solve Q10 Q.1) The general solution of the equation y* • v• ' -e' cos x , ÁN:a) y(x ) =c, +c; x +c,¢ ' Ae' cosa a Re' sin xb) y (x ) =c, +c,x +c,¢+ de' cos a + Be' sinxc) y(x) = Ce¯' + ei(Acosx + Bsinx) + xed) v (x ) -c, +c, x +c,e¨ +Ae ' cosx + Be ' sin xQ.2) The basis of the equation y'' + y" = 0 ,are:a) 1,1.eQ.3) Ihe radius of convergence of (1- x²)y" + 4y' + 6xy = 0 about xo = 0 , is:b) 1 , 1, e¯'c) 1,x,e²d) 1 , x , e"%3Da)+ ob) 2c) 0b) - 12e?* с) бе?d) 1d) -6e"Q.4) The Wronskian W(e~2", e", e²*), is:Q.5) The general solution of the equation y" + 9y" = 0 , is:a) y(x) = c; + cz + Acos2x + Bsin2xc) y(x) = c; + c; + Acos3x + Bsin3xQ.6) The singular point (s) of (x² + 4)y" – x y' + 4y = 0 , is (arc):a) 12Q.7) To write the power series Em=3 4m as the form £-zum ,we must shift m to become :a) m - 1Q.8) The characteristic equation for y"' + 3y" – 4y = 0 , is :a) 2' = 32?Q.9) The form of the particular solution of the equation y" – 9y = 4 sin 3x + 2cos3x , is:a) y,(x) = A cos3xc) y, (x) = A sin3xQ.10) The recurrence relation which obtained from solving y" + xy = 0 by using power series , is:a) 12e*b)y(x) = c, + c,x + Acos2x + Bsin2xd) y(x) = c, + c,x + Acos3x + Bsin3xb) ±2ic) 0d) (0 , ±2i }b) т - 2с) т +1d) m +2b) 23 = 322-4c) d* + 2 = -1d) 2' = -32? + 4%3Db) y,(x) = A cos3x + Bsin3xd) y, (x) = Axcos3x + Bxsin3xa) – am-1 = (m + 2)(m + 1)a,m+2b) am+2%3D(m +2)(m+1)с) а, 3D (т — 2)(m - 1)а,m+2d) am+2(m+2)(m+ 1)Q.11) Rewrite the series x E=1m am x™-1 + £m-0 amx™ as a single power series whose general terminvolves xm:a) E=s(am-1 – am-4) x™c) a, + Em-5(am-1- am-4) x™Q.12) The solution of y" + y' = 0 by using power series method , is:b) E-4(am-1 – am-3) x™d) a, + Em=1(m am + am) x™Am-4) xma) y(x) = ao – a;(1 –b) y(x) = a, + a(1+2!3!x*с) у (x) %3D а, + а,(х-:d)y(x) = ao + a,( x ++ ..)+5!2!4!(-1)™x"Q.13) The interval and radius of convergence for power series Em-ob) (-∞, +m), 1, respectively , are :d) (-m, +∞), 3m!a) (-∞, +∞), ∞c) (-∞, +∞), 2

Question

Solve Q10 Q.1) The general solution of the equation y* • v• ' -e' cos x , ÁN:a) y(x ) =c, +c; x +c,¢ ' Ae' cosa a Re' sin xb) y (x ) =c, +c,x +c,¢+ de' cos a + Be' sinxc) y(x) = Ce¯' + ei(Acosx + Bsinx) + xed) v (x ) -c, +c, x +c,e¨ +Ae ' cosx + Be ' sin xQ.2) The basis of the equation y'' + y&#34; = 0 ,are:a) 1,1.eQ.3) Ihe radius of convergence of (1- x²)y&#34; + 4y' + 6xy = 0 about xo = 0 , is:b) 1 , 1, e¯'c) 1,x,e²d) 1 , x , e&#34;%3Da)+ ob) 2c) 0b) - 12e?* с) бе?d) 1d) -6e&#34;Q.4) The Wronskian W(e~2&#34;, e&#34;, e²*), is:Q.5) The general solution of the equation y&#34; + 9y&#34; = 0 , is:a) y(x) = c; + cz + Acos2x + Bsin2xc) y(x) = c; + c; + Acos3x + Bsin3xQ.6) The singular point (s) of (x² + 4)y&#34; – x y' + 4y = 0 , is (arc):a) 12Q.7) To write the power series Em=3 4m as the form £-zum ,we must shift m to become :a) m - 1Q.8) The characteristic equation for y&#34;' + 3y&#34; – 4y = 0 , is :a) 2' = 32?Q.9) The form of the particular solution of the equation y&#34; – 9y = 4 sin 3x + 2cos3x , is:a) y,(x) = A cos3xc) y, (x) = A sin3xQ.10) The recurrence relation which obtained from solving y&#34; + xy = 0 by using power series , is:a) 12e*b)y(x) = c, + c,x + Acos2x + Bsin2xd) y(x) = c, + c,x + Acos3x + Bsin3xb) ±2ic) 0d) (0 , ±2i }b) т - 2с) т +1d) m +2b) 23 = 322-4c) d* + 2 = -1d) 2' = -32? + 4%3Db) y,(x) = A cos3x + Bsin3xd) y, (x) = Axcos3x + Bxsin3xa) – am-1 = (m + 2)(m + 1)a,m+2b) am+2%3D(m +2)(m+1)с) а, 3D (т — 2)(m - 1)а,m+2d) am+2(m+2)(m+ 1)Q.11) Rewrite the series x E=1m am x™-1 + £m-0 amx™ as a single power series whose general terminvolves xm:a) E=s(am-1 – am-4) x™c) a, + Em-5(am-1- am-4) x™Q.12) The solution of y&#34; + y' = 0 by using power series method , is:b) E-4(am-1 – am-3) x™d) a, + Em=1(m am + am) x™Am-4) xma) y(x) = ao – a;(1 –b) y(x) = a, + a(1+2!3!x*с) у (x) %3D а, + а,(х-:d)y(x) = ao + a,( x ++ ..)+5!2!4!(-1)™x&#34;Q.13) The interval and radius of convergence for power series Em-ob) (-∞, +m), 1, respectively , are :d) (-m, +∞), 3m!a) (-∞, +∞), ∞c) (-∞, +∞), 2

Accepted Answer

Answered: Image /qna-images/answer/d39524cd-8702-46ba-aaa9-147a18949145.jpg

- am-1 = (m + 2)(m + 1)a,m+2 %3D