ODE6.1

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y''-y=e^{{-t}},\,\,\,y(0)=1,y'(0)=0\,

{\mathcal  {L}}[y''-y]={\mathcal  {L}}[e^{{-t}}]\,

Let Y={\mathcal  {L}}[y]\,

{\mathcal  {L}}[y''-y]={\mathcal  {L}}[y'']-{\mathcal  {L}}[y]=s^{2}Y-sf(0)-f'(0)-Y\,

=s^{2}Y-s-Y={\frac  {1}{s+1}}\,

At this point all the information from the IC's are encoded into the transform. Solve for Y.

Y={\frac  {s^{2}+s+1}{(s+1)(s^{2}-1)}}\,


Find the partial fraction decomposition of {\frac  {s^{2}+s+1}{(s+1)(s^{2}-1)}}\,

{\frac  {s^{2}+s+1}{(s+1)(s^{2}-1)}}={\frac  {s^{2}+s+1}{(s+1)^{2}(s-1)}}={\frac  {A}{(s+1)^{2}}}+{\frac  {B}{s+1}}+{\frac  {C}{s-1}}\,

Multiply both sides by the common denominator.

s^{2}+s+1=A(s-1)+B(s+1)(s-1)+C(s+1)^{2}\,

s^{2}+s+1=s^{2}(B+C)+s(A+2C)+(-A-B+C)\,

Compare coefficients.

B+C=1,\,\,\,A+2C=1,\,\,\,-A-B+C=1\,

Solve this system of equations. The augmented matrix has equations as rows and variables as columns. The last column is the value of each equation.

{\begin{bmatrix}A&B&C&{}\\0&1&1&1\\1&0&2&1\\-1&-1&1&1\\\end{bmatrix}}={\begin{bmatrix}A&B&C&{}\\1&0&2&1\\-1&-1&1&1\\0&1&1&1\\\end{bmatrix}}={\begin{bmatrix}A&B&C&{}\\1&0&2&1\\0&-1&3&2\\0&1&1&1\\\end{bmatrix}}={\begin{bmatrix}A&B&C&{}\\1&0&2&1\\0&-1&3&2\\0&0&4&3\\\end{bmatrix}}\,

So C=3/4,\,\,\,B=1/4,\,\,\,A=-1/2\,

{\frac  {s^{2}+s+1}{(s+1)(s^{2}-1)}}={\frac  {-1/2}{(s+1)^{2}}}+{\frac  {1/4}{s+1}}+{\frac  {-1/2}{s-1}}\,


So Y={\frac  {-1/2}{(s+1)^{2}}}+{\frac  {1/4}{s+1}}+{\frac  {-1/2}{s-1}}\,

Use the tables here to find the inverse Laplace transforms.

y(t)={\mathcal  {L}}^{{-1}}[Y]={\frac  {-1}{2}}(te^{{-t}})+{\frac  {1}{4}}e^{{-t}}+{\frac  {3}{4}}e^{t}\,

Ordinary Differential Equations

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