2nd Order Ordinary Differential Equations

Number of Questions:	3
Marks Available:	11
Pass Percentage:	100%
Time Allowed:
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2nd Order Ordinary Differential Equations

Question 1

Method of undertermined coefficients

Solve:
\[\simplify[std]{(d^2y/dx^2)+{2*a}*(dy/dx)+{a^2+b^2}y}=0\]
which satisfies $y(0)=1$ and $y'(0)=\var{c}$ (where prime denotes the derivative).

Solution is:

$y=\;\;$interpreted as $\displaystyle{}$ Expected answer:interpreted as $\displaystyle{e^{ -4 x } \left ( \cos \left ( 7 x \right ) + \frac{ 6 }{ 7 } \sin \left ( 7 x \right ) \right )}$

Input all numbers as integers or fractions – not as decimals.

Score: 0/3

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Advice

The auxillary equation is $\simplify[std]{lambda^2+{2*a}lambda+{a^2+b^2}}=0$.

On solving this equation we get $\lambda=\simplify[std]{{-a}+{b}i}$ and $\lambda=\simplify[std]{{-a}-{b}i}$.

Hence the general solution is:
\[y = \simplify[std]{e^({-a}x)(A*sin({b}x)+B*cos({b}x))}\]
Note that
\[y'(x)=\simplify[std]{-{a}e^({-a}x)(A*sin({b}x)+B*cos({b}x))+e^({-a}x)({b}*A*cos({b}x)-{b}*B*sin({b}x))}\]
Using the conditions $y(0)=1$ and $y'(0)=\var{c}$ gives:
\[\begin{eqnarray*} B &=& 1\\ \simplify[std]{{b}A+{-a}B}&=& \var{c} \end{eqnarray*} \]
This gives $\displaystyle{A = \simplify[std]{{c+a}/{b}}}$.

Hence the solution is:

\[y=\simplify[std]{exp({- a} * x) * (cos({b} * x) + ({c+a} / {b}) * sin({b} * x))}\]

Question 2

Method of undetermined coefficients:

Find the solution of:
\[\simplify[std]{(d^2y/dx^2)+{2*a}*(dy/dx)+{a^2}y}=0\]
which satisfies $y(0)=\var{c}$ and $y(1)=\var{d}$.

0,0

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Solution is:

$y=\;\;$ interpreted as $\displaystyle{}$ Expected answer:interpreted as $\displaystyle{2 e^{ -4 x } + 325.589 x e^{ -4 x }}$

Input all numbers correct to 3 decimal places.

Your solution is graphed above once you have input the solution before submitting.

Note that the points $(0,\var{c}),\;(1,\var{d})$ which the curve has to go through are marked on the graph.

Score: 0/3

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Advice

The auxillary equation is $\simplify[std]{lambda^2+{2*a}lambda+{a^2}}=0$.

On solving this equation we get $\lambda=\var{-a}$ twice.

Hence the general solution is:
\[y = \simplify[std]{A*e^({-a}x)+B*x*e^({-a}x)}\]
The boundary conditions give:

$y(0)=\var{c} \Rightarrow A=\var{c}$

$y(1)=\var{d} \Rightarrow \simplify[std]{Ae^{-a}+Be^{-a}={d}}\Rightarrow A+B = \simplify[std1]{{d}e^{a}}$

So $B=\simplify[std1]{{d}e^{a}-{c}}=\var{f1}$ to 3 decimal places.

Hence the solution is:
\[y=\simplify[std1]{(({c} * Exp(({( - a)} * x))) + ({f1} * x * Exp(({( - a)} * x))))}\]

0,0

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Question 3

Find the complete general solution of the equation:
\[\simplify[std]{d^2y/dx^2+{2*p}*(dy/dx)+{p^2-q^2}y=x}\]
in the form $\displaystyle{Ae^{ax}+Be^{bx}+y_{PI}(x)}$ where $A$ and $B$ are arbitrary constants and $y_{PI}$ is a particular integral.

Note that we require that $a \gt b$.

a)

$a=\;\;$Expected answer:$\;\;\;b=\;\;$Expected answer:. Remember that we require $a \gt b$

Score: 0/2

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b)

$\displaystyle{y_{PI}=\;\;}$interpreted as $\displaystyle{}$ Expected answer:interpreted as $\displaystyle{\frac{ x }{ 9 } + \frac{ 10 }{ 81 }}$ .

Input all numbers as fractions or integers and not as decimals.

Score: 0/3

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Advice

First we find the Complementary Function (CF) i.e. the solution of:

\[\simplify[std]{d^2y/dx^2+{2*p}*(dy/dx)+{p^2-q^2}y=0}\]

The auxillary equation is $\simplify[std]{lambda^2+{2*p}lambda+{p^2-q^2}=0} \Rightarrow \simplify[std]{(lambda+{p-q})(lambda+{p+q})=0}\Rightarrow \lambda = \var{q-p}\textrm{ or }\lambda = \var{-q-p} $

Hence the CF is $\displaystyle{y_{CF}(x)=\simplify[std]{A*e^({q-p}x)+ B*e^({-q-p}x)}}$ for $A$, $B$ arbitrary constants.

So $a=\var{q-p}$ and $b=\var{-q-p}$ as we required $a \gt b$.

To find the Particular Integral (PI) for $\displaystyle{\simplify[std]{d^2y/dx^2+{2*p}*(dy/dx)+{p^2-q^2}y=x}}$, we observe that

$y_{PI}(x)=cx+d$ is a possible PI for constants $c,\;d$ to be found.

Substituting $y_{PI}(x)=cx+d$ in to the equation gives:

$\displaystyle \simplify[std]{{2*p}c+{p^2-q^2}*(cx+d)} =x \Rightarrow \simplify[std]{{p^2-q^2}cx+{p^2-q^2}d+{2*p}c}= x \Rightarrow c=\simplify[std]{1/{p^2-q^2}},\;\;d = \simplify[std]{{-2p}/{(p^2-q^2)^2}}$

So the particular integral is $\displaystyle{\simplify[std]{x/{p^2-q^2}-{2*p}/{(p^2-q^2)^2}}}$

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2nd Order Ordinary Differential Equations

2nd Order Ordinary Differential Equations

Advice

Advice

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b)

Advice