(D3-3D2+3D-1)y=(x+1)ex

1.	(D3-3D2+3D-1)y=(x+1)ex
Answer» SOLUTION TO SOLVE $\displaystyle\sf{ ({D}^{3} - 3 {D}^{2} + 3D - 1)y = (x + 1) {e}^{x} }$ EVALUATION Here the given differential equation is $\displaystyle\sf{ ({D}^{3} - 3 {D}^{2} + 3D - 1)y = (x + 1) {e}^{x} }$ Now complementary function is obtained by solving the differential equation $\displaystyle\sf{ ({D}^{3} - 3 {D}^{2} + 3D - 1)y = 0 }$ Let $\displaystyle\sf{ y = {e}^{mx} }$ be the trial solution Then the auxiliary equation is $\displaystyle\sf{ ({m}^{3} - 3 {m}^{2} + 3m - 1) = 0 }$ $\displaystyle\sf{ \implies \: {(m - 1)}^{3} = 0 }$ $\displaystyle\sf{ \implies \: m = 1,1,1}$ Thus three roots are REAL and equal So the complementary function is $\sf{y_{_{C.F}} = (a + bx + c {x}^{2} ) {e}^{x} }$ Now the particular solution is obtained as below $\displaystyle\sf{ y_{_{P.I}} = \frac{1}{({D}^{3} - 3 {D}^{2} + 3D - 1)} (x + 1) {e}^{x} }$ $\displaystyle\sf{ \implies \: y_{_{P.I}} = \frac{1}{{(D - 1)}^{3} } (x + 1) {e}^{x} }$ $\displaystyle\sf{ \implies \: y_{_{P.I}} = {e}^{x} \frac{1}{{(D + 1- 1)}^{3} } (x + 1) }$ $\displaystyle\sf{ \implies \: y_{_{P.I}} = {e}^{x} \frac{1}{{D }^{3} } (x + 1) }$ $\displaystyle\sf{ \implies \: y_{_{P.I}} = {e}^{x} \: \frac{ {(x + 1)}^{4} }{24} }$ $\displaystyle\sf{ \implies \: y_{_{P.I}} = \: \frac{ {(x + 1)}^{4} }{24} \: {e}^{x} }$ Hence the required solution is $\sf{y =y_{_{C.F}} + y_{_{P.I}}}$ $\displaystyle \sf{ \implies \: y = (a + bx + c {x}^{2} ) {e}^{x} + \frac{ {(x + 1)}^{4} }{24} \: {e}^{x} }$ FINAL ANSWER $\boxed{ \: \: \displaystyle \sf{ \: y = (a + bx + c {x}^{2} ) {e}^{x} + \frac{ {(x + 1)}^{4} }{24} \: {e}^{x} } \: \: \: }$ ━━━━━━━━━━━━━━━━ Learn more from Brainly :- 1. M+N(dy/dx)=0 where M and N are function of (A) x only (B) y only (C) constant (D) all of these brainly.in/question/38173299 2. This TYPE of equation is of the form dy/dx=f1(x,y)/f2(x,y) (A) variable seprable (B) homogeneous (C) exact (D) none ... brainly.in/question/38173619

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