Question

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


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Answer 1

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

$\boxed{ \: \: \displaystyle \sf{ \: y = (a + bx + c {x}^{2} ) {e}^{x} + \frac{ {(x + 1)}^{4} }{24} \: {e}^{x} } \: \: \: }$

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