Engineering Mathematics II • Digital Workbook • Page 21

Differential Equations Using Laplace Transforms

Initial-Value Problems and Partial Fractions

Differential equations using Laplace transforms are developed through three IOE past-paper questions. Transform each derivative, apply the initial conditions, resolve the resulting function and take the inverse transform.

UnitLaplace Transform
TopicDifferential Equations
Question Type4 Mark Problems

Theory Required for These Problems

Transform of Derivatives

\[L\{y'\}=sL\{y\}-y(0).\]

\[L\{y''\}=s^2L\{y\}-sy(0)-y'(0).\]
\[L\{y''\}\]\[=s^2L\{y\}-sy(0)-y'(0).\]
\[L\{y'''\}=s^3L\{y\}-s^2y(0)-sy'(0)-y''(0).\]
\[L\{y'''\}\]\[=s^3L\{y\}-s^2y(0)-sy'(0)-y''(0).\]
\[L\{y^{(4)}\}=s^4L\{y\}-s^3y(0)-s^2y'(0)-sy''(0)-y'''(0).\]
\[L\{y^{(4)}\}\]\[=s^4L\{y\}-s^3y(0)-s^2y'(0)\]\[{}-sy''(0)-y'''(0).\]

Initial Conditions

Write the complete transformed equation first. Then substitute the initial conditions without changing their signs.

Note: If \(y'(0)=-1\), then

\[-y'(0)=-(-1)=+1.\]

Partial Fractions

Factorise the denominator completely. Then write the partial-fraction form according to every factor and its order.

\[\frac{1}{(s+1)(s+2)^2}=\frac{A}{s+1}+\frac{B}{s+2}+\frac{C}{(s+2)^2}.\]
\[\frac{1}{(s+1)(s+2)^2}\]\[=\frac{A}{s+1}+\frac{B}{s+2}+\frac{C}{(s+2)^2}.\]

Useful Inverse Transforms

\[L^{-1}\left\{\frac1{s-a}\right\}=e^{at}.\]

\[L^{-1}\left\{\frac1{(s-a)^2}\right\}=te^{at}.\]

\[L^{-1}\left\{\frac1{(s-a)^3}\right\}=\frac{t^2}{2}e^{at}.\]

\[L^{-1}\left\{\frac{s-a}{(s-a)^2+b^2}\right\}=e^{at}\cos bt.\]
\[L^{-1}\left\{\frac{s-a}{(s-a)^2+b^2}\right\}\]\[=e^{at}\cos bt.\]
\[L^{-1}\left\{\frac{b}{(s-a)^2+b^2}\right\}=e^{at}\sin bt.\]
\[L^{-1}\left\{\frac{b}{(s-a)^2+b^2}\right\}\]\[=e^{at}\sin bt.\]
Method used: Write the given equation, take the Laplace transform on both sides, substitute the initial conditions, collect \(L\{x\}\) or \(L\{y\}\), write the inverse-transform form, resolve into partial fractions and then apply \(L^{-1}\).

Note on Partial Fractions

Simple Linear Factor\[(ax+b)\quad\Rightarrow\quad\frac{A}{ax+b}.\]
Linear Factor of Second Order
\[(ax+b)^2\quad\Rightarrow\quad\frac{A}{ax+b}+\frac{B}{(ax+b)^2}.\]
\[(ax+b)^2\quad\Rightarrow\]\[\frac{A}{ax+b}+\frac{B}{(ax+b)^2}.\]
Linear Factor of Third Order
\[(ax+b)^3\quad\Rightarrow\quad\frac{A}{ax+b}+\frac{B}{(ax+b)^2}+\frac{C}{(ax+b)^3}.\]
\[(ax+b)^3\quad\Rightarrow\]\[\frac{A}{ax+b}+\frac{B}{(ax+b)^2}\]\[{}+\frac{C}{(ax+b)^3}.\]
Irreducible Quadratic Factor
\[(ax^2+bx+c)\quad\Rightarrow\quad\frac{Ax+B}{ax^2+bx+c},\qquad b^2-4ac<0.\]
\[(ax^2+bx+c)\quad\Rightarrow\]\[\frac{Ax+B}{ax^2+bx+c},\qquad b^2-4ac<0.\]

Combined Form

\[\frac{N(x)}{(x-1)(x+2)^3(x^2+4)}=\frac{A}{x-1}+\frac{B}{x+2}+\frac{C}{(x+2)^2}+\frac{D}{(x+2)^3}+\frac{Ex+F}{x^2+4}.\]
\[\frac{N(x)}{(x-1)(x+2)^3(x^2+4)}\]\[=\frac{A}{x-1}+\frac{B}{x+2}\]\[{}+\frac{C}{(x+2)^2}+\frac{D}{(x+2)^3}\]\[{}+\frac{Ex+F}{x^2+4}.\]

Important: For a repeated factor, write one partial fraction for every power from (1) up to its highest power. For an irreducible quadratic factor, the numerator must be linear.

Note: In Laplace-transform problems, replace \(x\) by \(s\).

Continue through the IOE Engineering Mathematics II Digital Workbook. This differential equations using laplace transforms page follows the Engineering Mathematics II course sequence in the Tribhuvan University curriculum.
45

Initial-Value Problem with a Repeated Factor

4 Marks
PAST-PAPER REFERENCE: TU IOE • 2083 Baisakh • Back (New Course) • ENSH 151 • Q. 12 • 4 Marks

Using the Laplace transform method, solve

\[y''(t)+4y'(t)+4y(t)=e^{-t},\]

subject to \(y(0)=0,\ y'(0)=0\).

Solution

Laplace transform and partial fractions

Given equation is

\[y''+4y'+4y=e^{-t}.\]

Taking Laplace transform on both sides,

\[L\{y''+4y'+4y\}=L\{e^{-t}\}.\]

i.e.

\[L\{y''\}+4L\{y'\}+4L\{y\}=\frac1{s+1}.\]

Using the transform of derivatives,

\[s^2L\{y\}-sy(0)-y'(0)+4[sL\{y\}-y(0)]+4L\{y\}=\frac1{s+1}.\]
\[s^2L\{y\}-sy(0)-y'(0)\]\[{}+4[sL\{y\}-y(0)]+4L\{y\}\]\[=\frac1{s+1}.\]

Using \(y(0)=0\) and \(y'(0)=0\),

\[s^2L\{y\}+4sL\{y\}+4L\{y\}=\frac1{s+1}.\]
\[s^2L\{y\}+4sL\{y\}+4L\{y\}\]\[=\frac1{s+1}.\]

Therefore,

\[L\{y\}(s^2+4s+4)=\frac1{s+1}.\]

Since \(s^2+4s+4=(s+2)^2\),

\[L\{y\}(s+2)^2=\frac1{s+1}.\]

Hence,

\[L\{y\}=\frac1{(s+1)(s+2)^2}.\]

Therefore,

\[y=L^{-1}\left\{\frac1{(s+1)(s+2)^2}\right\}\qquad(i)\]

Now, resolving into partial fractions, let

\[\frac1{(s+1)(s+2)^2}=\frac{A}{s+1}+\frac{B}{s+2}+\frac{C}{(s+2)^2}.\]
\[\frac1{(s+1)(s+2)^2}\]\[=\frac{A}{s+1}+\frac{B}{s+2}+\frac{C}{(s+2)^2}.\]

Multiplying by \((s+1)(s+2)^2\),

\[1=A(s+2)^2+B(s+1)(s+2)+C(s+1).\]
\[1=A(s+2)^2+B(s+1)(s+2)\]\[{}+C(s+1).\]

Putting \(s=-1\),

\[1=A(-1+2)^2\]\[A=1.\]

Putting \(s=-2\),

\[1=C(-2+1)\]\[C=-1.\]

Comparing the coefficients of \(s^2\),

\[0=A+B.\]

Therefore,

\[0=1+B\]\[B=-1.\]

Hence,

\[\frac1{(s+1)(s+2)^2}=\frac1{s+1}-\frac1{s+2}-\frac1{(s+2)^2}.\]
\[\frac1{(s+1)(s+2)^2}\]\[=\frac1{s+1}-\frac1{s+2}-\frac1{(s+2)^2}.\]

By \((i)\),

\[y=L^{-1}\left\{\frac1{s+1}-\frac1{s+2}-\frac1{(s+2)^2}\right\}.\]
\[y=L^{-1}\left\{\frac1{s+1}-\frac1{s+2}\right.\]\[\left.{}-\frac1{(s+2)^2}\right\}.\]

Therefore,

\[y=e^{-t}-e^{-2t}-te^{-2t}.\]

Hence,

\[y=e^{-t}-(1+t)e^{-2t}.\]
\[y=e^{-t}-(1+t)e^{-2t}\]
46

Initial-Value Problem with Distinct Linear Factors

4 Marks
PAST-PAPER REFERENCE: TU IOE • 2082 Bhadra • Regular (New Course) • ENSH 151 • Q. 12 • 4 Marks

Using the Laplace transform, solve the initial-value problem

\[x''-3x'+2x=e^{-t},\]

where \(x(0)=0,\ x'(0)=1\).

Solution

Laplace transform and partial fractions

Given equation is

\[x''-3x'+2x=e^{-t}.\]

Taking Laplace transform on both sides,

\[L\{x''-3x'+2x\}=L\{e^{-t}\}.\]

i.e.

\[L\{x''\}-3L\{x'\}+2L\{x\}=\frac1{s+1}.\]

Using the transform of derivatives,

\[s^2L\{x\}-sx(0)-x'(0)-3[sL\{x\}-x(0)]+2L\{x\}=\frac1{s+1}.\]
\[s^2L\{x\}-sx(0)-x'(0)\]\[{}-3[sL\{x\}-x(0)]+2L\{x\}\]\[=\frac1{s+1}.\]

Using \(x(0)=0\) and \(x'(0)=1\),

\[s^2L\{x\}-1-3sL\{x\}+2L\{x\}=\frac1{s+1}.\]
\[s^2L\{x\}-1-3sL\{x\}+2L\{x\}\]\[=\frac1{s+1}.\]

Therefore,

\[L\{x\}(s^2-3s+2)-1=\frac1{s+1}.\]

Hence,

\[L\{x\}(s^2-3s+2)=1+\frac1{s+1}.\]

Taking the L.C.M.,

\[L\{x\}(s^2-3s+2)=\frac{s+1+1}{s+1}\]\[=\frac{s+2}{s+1}.\]

Since \(s^2-3s+2=(s-1)(s-2)\),

\[L\{x\}(s-1)(s-2)=\frac{s+2}{s+1}.\]

Therefore,

\[L\{x\}=\frac{s+2}{(s+1)(s-1)(s-2)}.\]
\[L\{x\}\]\[=\frac{s+2}{(s+1)(s-1)(s-2)}.\]

Hence,

\[x=L^{-1}\left\{\frac{s+2}{(s+1)(s-1)(s-2)}\right\}\qquad(i)\]
\[x=L^{-1}\left\{\frac{s+2}{(s+1)(s-1)(s-2)}\right\}\]\[\qquad(i)\]

Now, resolving into partial fractions, let

\[\frac{s+2}{(s+1)(s-1)(s-2)}=\frac{A}{s+1}+\frac{B}{s-1}+\frac{C}{s-2}.\]
\[\frac{s+2}{(s+1)(s-1)(s-2)}\]\[=\frac{A}{s+1}+\frac{B}{s-1}+\frac{C}{s-2}.\]

Multiplying by \((s+1)(s-1)(s-2)\),

\[s+2=A(s-1)(s-2)+B(s+1)(s-2)+C(s+1)(s-1).\]
\[s+2=A(s-1)(s-2)\]\[{}+B(s+1)(s-2)+C(s+1)(s-1).\]

Putting \(s=-1\),

\[1=A(-2)(-3)=6A\]\[A=\frac16.\]

Putting \(s=1\),

\[3=B(2)(-1)=-2B\]\[B=-\frac32.\]

Putting \(s=2\),

\[4=C(3)(1)=3C\]\[C=\frac43.\]

Hence,

\[\frac{s+2}{(s+1)(s-1)(s-2)}=\frac1{6(s+1)}-\frac3{2(s-1)}+\frac4{3(s-2)}.\]
\[\frac{s+2}{(s+1)(s-1)(s-2)}\]\[=\frac1{6(s+1)}-\frac3{2(s-1)}\]\[{}+\frac4{3(s-2)}.\]

By \((i)\),

\[x=L^{-1}\left\{\frac1{6(s+1)}-\frac3{2(s-1)}+\frac4{3(s-2)}\right\}.\]
\[x=L^{-1}\left\{\frac1{6(s+1)}-\frac3{2(s-1)}\right.\]\[\left.{}+\frac4{3(s-2)}\right\}.\]

Therefore,

\[x=\frac16e^{-t}-\frac32e^t+\frac43e^{2t}.\]
\[x=\frac16e^{-t}-\frac32e^t\]\[{}+\frac43e^{2t}.\]
\[x=\frac16e^{-t}-\frac32e^t+\frac43e^{2t}.\]
\[x=\frac16e^{-t}-\frac32e^t\]\[{}+\frac43e^{2t}.\]
47

Initial-Value Problem with a Quadratic Factor

4 Marks
PAST-PAPER REFERENCE: TU IOE • 2081 Ashwin • Regular (New Course, 2080 Batch) • SH 151 • Q. 12 • 4 Marks

Using the Laplace transform technique, solve

\[y''(t)-y'(t)+6y(t)=e^{-t},\]

subject to \(y(0)=0,\ y'(0)=0\).

Solution

Partial fractions and shifted transforms

Given equation is

\[y''-y'+6y=e^{-t}.\]

Taking Laplace transform on both sides,

\[L\{y''-y'+6y\}=L\{e^{-t}\}.\]

i.e.

\[L\{y''\}-L\{y'\}+6L\{y\}=\frac1{s+1}.\]

Using the transform of derivatives,

\[s^2L\{y\}-sy(0)-y'(0)-[sL\{y\}-y(0)]+6L\{y\}=\frac1{s+1}.\]
\[s^2L\{y\}-sy(0)-y'(0)\]\[{}-[sL\{y\}-y(0)]+6L\{y\}\]\[=\frac1{s+1}.\]

Using \(y(0)=0\) and \(y'(0)=0\),

\[s^2L\{y\}-sL\{y\}+6L\{y\}=\frac1{s+1}.\]
\[s^2L\{y\}-sL\{y\}+6L\{y\}\]\[=\frac1{s+1}.\]

Therefore,

\[L\{y\}(s^2-s+6)=\frac1{s+1}.\]

Hence,

\[L\{y\}=\frac1{(s+1)(s^2-s+6)}.\]

Therefore,

\[y=L^{-1}\left\{\frac1{(s+1)(s^2-s+6)}\right\}\qquad(i)\]
\[y=L^{-1}\left\{\frac1{(s+1)(s^2-s+6)}\right\}\]\[\qquad(i)\]

Now, resolving into partial fractions, let

\[\frac1{(s+1)(s^2-s+6)}=\frac{A}{s+1}+\frac{Bs+C}{s^2-s+6}.\]
\[\frac1{(s+1)(s^2-s+6)}\]\[=\frac{A}{s+1}+\frac{Bs+C}{s^2-s+6}.\]

Multiplying by \((s+1)(s^2-s+6)\),

\[1=A(s^2-s+6)+(Bs+C)(s+1).\]
\[1=A(s^2-s+6)+(Bs+C)(s+1).\]

Expanding,

\[1=As^2-As+6A+Bs^2+Bs+Cs+C.\]
\[1=As^2-As+6A+Bs^2+Bs+Cs+C.\]

Collecting like powers of \(s\),

\[1=(A+B)s^2+(-A+B+C)s+(6A+C).\]
\[1=(A+B)s^2+(-A+B+C)s\]\[{}+(6A+C).\]

Comparing the coefficients of \(s^2\), \(s\) and the constant terms,

\[A+B=0\qquad(1)\]\[-A+B+C=0\qquad(2)\]\[6A+C=1\qquad(3)\]

From \((1)\),

\[B=-A.\]

Substituting \(B=-A\) in \((2)\),

\[-A-A+C=0\]\[C=2A.\]

Substituting \(C=2A\) in \((3)\),

\[6A+2A=1\]\[8A=1\]\[A=\frac18.\]

Therefore,

\[B=-\frac18,\qquad C=\frac14.\]

Hence,

\[\frac1{(s+1)(s^2-s+6)}=\frac1{8(s+1)}+\frac{-\frac{s}{8}+\frac14}{s^2-s+6}.\]
\[\frac1{(s+1)(s^2-s+6)}\]\[=\frac1{8(s+1)}+\frac{-\frac{s}{8}+\frac14}{s^2-s+6}.\]

Completing the square in the quadratic factor,

\[s^2-s+6=s^2-s+\frac14+6-\frac14\]\[=\left(s-\frac12\right)^2+\frac{23}{4}\]\[=\left(s-\frac12\right)^2+\left(\frac{\sqrt{23}}2\right)^2.\]

Also,

\[-\frac{s}{8}+\frac14=-\frac18\left(s-\frac12\right)+\frac3{16}.\]

Therefore,

\[\frac1{(s+1)(s^2-s+6)}=\frac1{8(s+1)}-\frac18\frac{s-\frac12}{\left(s-\frac12\right)^2+\left(\frac{\sqrt{23}}2\right)^2}+\frac3{8\sqrt{23}}\frac{\frac{\sqrt{23}}2}{\left(s-\frac12\right)^2+\left(\frac{\sqrt{23}}2\right)^2}.\]
\[\frac1{(s+1)(s^2-s+6)}=\frac1{8(s+1)}\]\[{}-\frac18\frac{s-\frac12}{\left(s-\frac12\right)^2+\left(\frac{\sqrt{23}}2\right)^2}\]\[{}+\frac3{8\sqrt{23}}\frac{\frac{\sqrt{23}}2}{\left(s-\frac12\right)^2+\left(\frac{\sqrt{23}}2\right)^2}.\]

By \((i)\),

\[y=L^{-1}\left\{\frac1{8(s+1)}-\frac18\frac{s-\frac12}{\left(s-\frac12\right)^2+\left(\frac{\sqrt{23}}2\right)^2}+\frac3{8\sqrt{23}}\frac{\frac{\sqrt{23}}2}{\left(s-\frac12\right)^2+\left(\frac{\sqrt{23}}2\right)^2}\right\}.\]
\[y=L^{-1}\left\{\frac1{8(s+1)}\right.\]\[\left.{}-\frac18\frac{s-\frac12}{\left(s-\frac12\right)^2+\left(\frac{\sqrt{23}}2\right)^2}\right.\]\[\left.{}+\frac3{8\sqrt{23}}\frac{\frac{\sqrt{23}}2}{\left(s-\frac12\right)^2+\left(\frac{\sqrt{23}}2\right)^2}\right\}.\]

Taking the inverse Laplace transform,

\[y=\frac18e^{-t}-\frac18e^{t/2}\cos\frac{\sqrt{23}t}{2}+\frac3{8\sqrt{23}}e^{t/2}\sin\frac{\sqrt{23}t}{2}.\]
\[y=\frac18e^{-t}\]\[{}-\frac18e^{t/2}\cos\frac{\sqrt{23}t}{2}\]\[{}+\frac3{8\sqrt{23}}e^{t/2}\sin\frac{\sqrt{23}t}{2}.\]
\[y=\frac18e^{-t}-\frac18e^{t/2}\cos\frac{\sqrt{23}t}{2}+\frac3{8\sqrt{23}}e^{t/2}\sin\frac{\sqrt{23}t}{2}.\]
\[y=\frac18e^{-t}\]\[{}-\frac18e^{t/2}\cos\frac{\sqrt{23}t}{2}\]\[{}+\frac3{8\sqrt{23}}e^{t/2}\sin\frac{\sqrt{23}t}{2}.\]

A Few Important Questions

Differential Equation with a Repeated Factor of Third Order

Solve by using the Laplace transform:

\[x''-2x'+x=e^t,\]

where \(x(0)=2\) and \(x'(0)=-1\).

Show Solution

Given equation is

\[x''-2x'+x=e^t.\]

Taking Laplace transform on both sides,

\[L\{x''-2x'+x\}=L\{e^t\}.\]

i.e.

\[L\{x''\}-2L\{x'\}+L\{x\}=\frac1{s-1}.\]

Using the transform of derivatives,

\[s^2L\{x\}-sx(0)-x'(0)-2[sL\{x\}-x(0)]+L\{x\}=\frac1{s-1}.\]
\[s^2L\{x\}-sx(0)-x'(0)\]\[{}-2[sL\{x\}-x(0)]+L\{x\}\]\[=\frac1{s-1}.\]

Using \(x(0)=2\) and \(x'(0)=-1\),

\[s^2L\{x\}-2s+1-2[sL\{x\}-2]+L\{x\}=\frac1{s-1}.\]
\[s^2L\{x\}-2s+1\]\[{}-2[sL\{x\}-2]+L\{x\}\]\[=\frac1{s-1}.\]

Opening the bracket,

\[s^2L\{x\}-2s+1-2sL\{x\}+4+L\{x\}=\frac1{s-1}.\]
\[s^2L\{x\}-2s+1-2sL\{x\}\]\[{}+4+L\{x\}=\frac1{s-1}.\]

Collecting \(L\{x\}\),

\[L\{x\}(s^2-2s+1)-2s+5=\frac1{s-1}.\]
\[L\{x\}(s^2-2s+1)-2s+5\]\[=\frac1{s-1}.\]

Therefore,

\[L\{x\}(s^2-2s+1)=\frac1{s-1}+2s-5.\]

Taking the L.C.M.,

\[L\{x\}(s^2-2s+1)=\frac{1+(2s-5)(s-1)}{s-1}.\]
\[L\{x\}(s^2-2s+1)\]\[=\frac{1+(2s-5)(s-1)}{s-1}.\]

Expanding the numerator,

\[L\{x\}(s^2-2s+1)=\frac{1+2s^2-2s-5s+5}{s-1}\]\[=\frac{2s^2-7s+6}{s-1}.\]

Since \(s^2-2s+1=(s-1)^2\),

\[L\{x\}(s-1)^2=\frac{2s^2-7s+6}{s-1}.\]

Hence,

\[L\{x\}=\frac{2s^2-7s+6}{(s-1)^3}.\]

Therefore,

\[x=L^{-1}\left\{\frac{2s^2-7s+6}{(s-1)^3}\right\}\qquad(i)\]

Now, resolving into partial fractions, let

\[\frac{2s^2-7s+6}{(s-1)^3}=\frac{A}{s-1}+\frac{B}{(s-1)^2}+\frac{C}{(s-1)^3}.\]
\[\frac{2s^2-7s+6}{(s-1)^3}\]\[=\frac{A}{s-1}+\frac{B}{(s-1)^2}+\frac{C}{(s-1)^3}.\]

Multiplying by \((s-1)^3\),

\[2s^2-7s+6=A(s-1)^2+B(s-1)+C.\]

Putting \(s=1\),

\[2-7+6=C\]\[C=1.\]

Comparing the coefficients of \(s^2\),

\[2=A.\]

Therefore,

\[A=2.\]

Putting \(s=0\),

\[6=A-B+C.\]

Substituting \(A=2\) and \(C=1\),

\[6=2-B+1\]\[B=-3.\]

Hence,

\[\frac{2s^2-7s+6}{(s-1)^3}=\frac2{s-1}-\frac3{(s-1)^2}+\frac1{(s-1)^3}.\]
\[\frac{2s^2-7s+6}{(s-1)^3}\]\[=\frac2{s-1}-\frac3{(s-1)^2}+\frac1{(s-1)^3}.\]

By \((i)\),

\[x=L^{-1}\left\{\frac2{s-1}-\frac3{(s-1)^2}+\frac1{(s-1)^3}\right\}.\]
\[x=L^{-1}\left\{\frac2{s-1}-\frac3{(s-1)^2}\right.\]\[\left.{}+\frac1{(s-1)^3}\right\}.\]

Taking the inverse Laplace transform,

\[x=2e^t-3te^t+\frac{t^2}{2}e^t.\]
\[x=2e^t-3te^t+\frac{t^2}{2}e^t\]
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