Q23E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Let\({\bf{L}}\left[{\bf{y}}\right]{\bf{=y'''+y'+ xy,}}\,\,\,\,\,\,\,{{\bf{y}}_{\bf{1}}}\left({\bf{x}}\right){\bf{=sinx,}}\)and\({{\bf{y}}_{\bf{2}}}\left({\bf{x}}\right){\bf{=x}}\).Verifythat\({\bf{L}}\left[{{{\bf{y}}_{\bf{1}}}}\right]\left( {\bf{x}} \right){\bf{=xsinx,}}\)and\({\bf{L}}\left[ {{{\bf{y}}_{\bf{2}}}} \right]\left( {\bf{x}} \right){\bf{ = }}{{\bf{x}}^{\bf{2}}}{\bf{ + 1}}\). Then use the superposition principle (linearity) to find a solution to the differential equation: 

 

(a) \({\bf{L}}\left[ {\bf{y}} \right]{\bf{ = 2xsinx - }}{{\bf{x}}^{\bf{2}}}{\bf{ - 1}}\)

(b) \({\bf{L}}\left[ {\bf{y}} \right]{\bf{ = 4}}{{\bf{x}}^{\bf{2}}}{\bf{ + 4 - 6xsinx}}\)

Step-by-Step Solution

Answer

(a) Thus,itisverifiedthat\({\bf{L}}\left[{{{\bf{y}}_{\bf{1}}}}\right]\left({\bf{x}}\right){\bf{ = xsinx}},\) and the solution to the differential equation is\({\bf{y = 2sinx -x}}.\)

 

 (b) Thus, it is verified that\({\bf{L}}\left[{{{\bf{y}}_{\bf{2}}}}\right]\left({\bf{x}} \right){\bf{=}}{{\bf{x}}^{\bf{2}}}{\bf{+1}},\)and the solution to the differential equation is\({\bf{y = 4x - 6sinx}}.\)

 

Step1:Firstlyverifythat \({\bf{L}}\left[{{{\bf{y}}_{\bf{1}}}}\right]\left({\bf{x}}\right){\bf{ = xsinx}}\) ,

 

The given equation is,

 

\(\begin{array}{c}{\bf{L}}\left[{\bf{y}}\right]{\bf{=y'''+y'+xy,}}\,\,\,\,\,\,\,.......\left({\bf{1}}\right)\\{{\bf{y}}_{\bf{1}}}\left({\bf{x}}\right){\bf{=sinx,}}\end{array}\)

 

Solve for\({{\bf{y}}_{\bf{1}}}\),

 

\(\begin{array}{c}{{\bf{y}}_{\bf{1}}}\left({\bf{x}}\right){\bf{=sinx}}\\{{\bf{y}}_{\bf{1}}}{\bf{'}}\left({\bf{x}}\right){\bf{=cosx}}\\{{\bf{y}}_{\bf{1}}}{\bf{''}}\left({\bf{x}}\right){\bf{=-sinx}}\\{{\bf{y}}_{\bf{1}}}{\bf{'''}}\left({\bf{x}}\right){\bf{=cosx}}\end{array}\)

 

Replacing \({\bf{y}}\) with \({{\bf{y}}_{\bf{1}}}\)in the equation (1),

 

\(\begin{array}{c}{\bf{L}}\left[{{{\bf{y}}_{\bf{1}}}}\right]{\bf{=}}{{\bf{y}}_{\bf{1}}}{\bf{'''+}}{{\bf{y}}_{\bf{1}}}{\bf{'+x}}{{\bf{y}}_{\bf{1}}}\\{\bf{=-cosx+cosx+ xsinx}}\\{\bf{ = xsinx}}\end{array}\)

 

Thus, it is verified that \({\bf{L}}\left[{{{\bf{y}}_{\bf{1}}}}\right]\left({\bf{x}} \right){\bf{=xsinx}}.\)

Step 2: Now verify that \({\bf{L}}\left[{{{\bf{y}}_{\bf{2}}}}\right]\left({\bf{x}} \right){\bf{=}}{{\bf{x}}^{\bf{2}}}{\bf{+1}}\) ,

Given,

 

\({{\bf{y}}_{\bf{2}}}\left( {\bf{x}} \right){\bf{ = x}}\)

 

Solvefor\({{\bf{y}}_{\bf{2}}}\),\(\begin{array}{c}{{\bf{y}}_{\bf{2}}}\left({\bf{x}}\right){\bf{=x}}\\{{\bf{y}}_{\bf{2}}}{\bf{'}}\left({\bf{x}}\right){\bf{=1}}\\{{\bf{y}}_{\bf{2}}}{\bf{''}}\left({\bf{x}}\right){\bf{=0}}\\{{\bf{y}}_{\bf{2}}}{\bf{'''}}\left({\bf{x}}\right){\bf{=0}}\end{array}\)

 

Replacing \({\bf{y}}\) with \({{\bf{y}}_{\bf{2}}}\) in the equation (1),

 

\(\begin{array}{c}{\bf{L}}\left[{{{\bf{y}}_{\bf{2}}}}\right]{\bf{=}}{{\bf{y}}_{\bf{2}}}{\bf{'''+}}{{\bf{y}}_{\bf{2}}}{\bf{'+x}}{{\bf{y}}_{\bf{2}}}\\{\bf{=0+1+x}}\left({\bf{x}}\right)\\{\bf{=}}{{\bf{x}}^{\bf{2}}}{\bf{+1}}\end{array}\)

 

Thus, it is verified that \({\bf{L}}\left[{{{\bf{y}}_{\bf{2}}}}\right]\left({\bf{x}} \right){\bf{=}}{{\bf{x}}^{\bf{2}}}{\bf{+1}}\)

Step 3: Use the superposition principle (linearity) for part (a);

If \({\bf{A}}{{\bf{x}}_{\bf{1}}}{\bf{=}}{{\bf{f}}_{\bf{1}}}{\bf{,}}\,\,{\bf{A}}{{\bf{x}}_{\bf{2}}}{\bf{ = }}{{\bf{f}}_{\bf{2}}}\)

 

Then \({\bf{A}}\left({{{\bf{a}}_{\bf{1}}}{{\bf{x}}_{\bf{1}}}{\bf{+}}{{\bf{a}}_{\bf{2}}}{{\bf{x}}_{\bf{2}}}}\right){\bf{=}}{{\bf{a}}_{\bf{1}}}{{\bf{f}}_{\bf{1}}}{\bf{+}}{{\bf{a}}_{\bf{2}}}{{\bf{f}}_{\bf{2}}}\)

 

Given equation,

 

\(\begin{array}{c}{\bf{L}}\left[{\bf{y}}\right]{\bf{=2xsinx-}}{{\bf{x}}^{\bf{2}}}{\bf{-1}}\\{\bf{L}}\left[{\bf{y}}\right]{\bf{=2}}\left({{\bf{xsinx}}}\right){\bf{1}}\left({{{\bf{x}}^{\bf{2}}}{\bf{+1}}}\right)\end{array}\)

 

Use the value of \({\bf{L}}\left[{{{\bf{y}}_{\bf{1}}}}\right]\left({\bf{x}}\right){\bf{= xsinx}}\) and \({\bf{L}}\left[{{{\bf{y}}_{\bf{2}}}}\right]\left({\bf{x}}\right){\bf{=}}{{\bf{x}}^{\bf{2}}}{\bf{ + 1}}\).

 

\({\bf{L}}\left[{\bf{y}}\right]{\bf{ = 2}}\left( {{\bf{L}}\left[ {{{\bf{y}}_{\bf{1}}}} \right]} \right){\bf{-1}}\left({{\bf{L}}\left[{{{\bf{y}}_{\bf{2}}}} \right]}\right)\)

 

Using superposition,

 

\(\begin{array}{c}{\bf{L}}\left[{\bf{y}}\right]{\bf{=L}}\left[{{\bf{2}}{{\bf{y}}_{\bf{1}}}{\bf{}}{{\bf{y}}_{\bf{2}}}}\right]\\{\bf{y=2}}{{\bf{y}}_{\bf{1}}}{\bf{}}{{\bf{y}}_{\bf{2}}}\end{array}\)

 

Substitute the value of \({{\bf{y}}{\bf{1}}}\left({\bf{x}}\right){\bf{=sinx}}\) and \({{\bf{y}}_{\bf{2}}}\left({\bf{x}}\right){\bf{=x}}\).

 

\({\bf{y=2sinx-x}}\)

 

Thus, the solution to the differential equation is\({\bf{y=2sinx-x}}.\)

Step 4: Use the superposition principle (linearity) for part (b);

If \({\bf{A}}{{\bf{x}}_{\bf{1}}}{\bf{=}}{{\bf{f}}_{\bf{1}}}{\bf{,}}\,\,{\bf{A}}{{\bf{x}}_{\bf{2}}}{\bf{ = }}{{\bf{f}}_{\bf{2}}}\)

 

Then \({\bf{A}}\left({{{\bf{a}}_{\bf{1}}}{{\bf{x}}_{\bf{1}}}{\bf{+ }}{{\bf{a}}_{\bf{2}}}{{\bf{x}}_{\bf{2}}}}\right){\bf{=}}{{\bf{a}}_{\bf{1}}}{{\bf{f}}_{\bf{1}}}{\bf{+}}{{\bf{a}}_{\bf{2}}}{{\bf{f}}_{\bf{2}}}\)

 

Given equation,

 

\(\begin{array}{c}{\bf{L}}\left[{\bf{y}}\right]{\bf{=4}}{{\bf{x}}^{\bf{2}}}{\bf{+4- 6xsinx}}\\{\bf{L}}\left[{\bf{y}}\right]{\bf{=4}}\left({{{\bf{x}}^{\bf{2}}}{\bf{+1}}}\right){\bf{-6}}\left({{\bf{xsinx}}}\right)\end{array}\)

 

Use the value of \({\bf{L}}\left[{{{\bf{y}}_{\bf{1}}}}\right]\left({\bf{x}}\right){\bf{= xsinx}}\) and\({\bf{L}}\left[{{{\bf{y}}_{\bf{2}}}}\right]\left({\bf{x}}\right){\bf{= }}{{\bf{x}}^{\bf{2}}}{\bf{ + 1}}\).

 

\({\bf{L}}\left[{\bf{y}}\right]{\bf{=4}}\left({{\bf{L}}\left[{{{\bf{y}}_{\bf{2}}}}\right]} \right){\bf{-6}}\left({{\bf{L}}\left[{{{\bf{y}}_{\bf{1}}}}\right]}\right)\)

 

Using superposition,

 

\(\begin{array}{c}{\bf{L}}\left[{\bf{y}}\right]{\bf{=L}}\left[{{\bf{4}}{{\bf{y}}_{\bf{2}}}{\bf{6}}{{\bf{y}}_{\bf{1}}}}\right]\\{\bf{y=4}}{{\bf{y}}_{\bf{2}}}{\bf{6}}{{\bf{y}}_{\bf{1}}}\end{array}\)

 

Substitute the value of \({{\bf{y}}{\bf{1}}}\left({\bf{x}}\right){\bf{=sinx}}\) and \({{\bf{y}}_{\bf{2}}}\left({\bf{x}}\right){\bf{=x}}\).

 

\({\bf{y = 4x - 6sinx}}\)

 

Thus, the solution to the differential equation is\({\bf{y=4x-6sinx}}.\)