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Obtain the general solution of the differential equation $$\frac{d^{2}y}{dx^{2}} + 5 \frac{dy}{dx} + 4y = 10 \sin 3x - 20 \cos 3x.$$ Show that, for large positive $x$ and independently of the initial conditions, $y \approx R \sin(3x + \phi)$, where the constants $R$ and $\phi$, such that $R > 0$ and $0 < \phi < 2\pi$, are to be determined correct to 2 decimal places. - CIE - A-Level Further Maths - Question 8 - 2010 - Paper 1

Question 8

$Obtain-the-general-solution-of-the-differential-equation-$$\frac{d^{2}y}{dx^{2}}-+-5-\frac{dy}{dx}-+-4y-=-10-\sin-3x---20-\cos-3x.$$---Show-that,-for-large-positive-$x$-and-independently-of-the-initial-conditions,---$y-\approx-R-\sin(3x-+-\phi)$,---where-the-constants-$R$-and-$\phi$,-such-that-$R->-0$-and-$0-<-\phi-<-2\pi$,-are-to-be-determined-correct-to-2-decimal-places.-CIE-A-Level Further Maths-Question 8-2010-Paper 1.png$

Obtain the general solution of the differential equation $$\frac{d^{2}y}{dx^{2}} + 5 \frac{dy}{dx} + 4y = 10 \sin 3x - 20 \cos 3x.$$ Show that, for large positive ... show full transcript

Worked Solution & Example Answer:Obtain the general solution of the differential equation $$\frac{d^{2}y}{dx^{2}} + 5 \frac{dy}{dx} + 4y = 10 \sin 3x - 20 \cos 3x.$$ Show that, for large positive $x$ and independently of the initial conditions, $y \approx R \sin(3x + \phi)$, where the constants $R$ and $\phi$, such that $R > 0$ and $0 < \phi < 2\pi$, are to be determined correct to 2 decimal places. - CIE - A-Level Further Maths - Question 8 - 2010 - Paper 1

Step 1

Obtain the complementary function

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Answer

The complementary function for the differential equation can be found by solving the characteristic equation associated with the homogeneous part, which is given as:

$r^{2} + 5r + 4 = 0$

Factoring gives us:

$(r + 1)(r + 4) = 0$

Thus, the roots are $r = -1$ and $r = -4$ . Therefore, the complementary function is:

$y_c = Ae^{-x} + Be^{-4x}$

Step 2

Find a particular integral

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Answer

To find a particular integral, we can use the method of undetermined coefficients. We assume a particular solution of the form:

$y_p = P \sin 3x + Q \cos 3x$

Substituting this into the differential equation and equating coefficients will allow us to solve for $P$ and $Q$ .

After substitution, we find:

$-5P + 15Q = 10$
$15P - 5Q = -20$

Solving these equations gives:

$P = -\frac{7}{5}, \, Q = -\frac{1}{5}$

Step 3

Combine the solutions

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Answer

The general solution of the differential equation is formed by combining the complementary and particular solutions:

$y = Ae^{-x} + Be^{-4x} - \frac{7}{5} \sin 3x - \frac{1}{5} \cos 3x$

Step 4

Analyze for large $x$

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Answer

For large positive $x$ , the terms $Ae^{-x}$ and $Be^{-4x}$ approach zero. Therefore, we focus on:

$y \approx -\frac{7}{5} \sin 3x - \frac{1}{5} \cos 3x$

This can be rewritten using amplitude and phase shift:

$y \approx R \sin(3x + \phi)$

where $R$ is the amplitude and $\phi$ is the phase shift. We find:

$R = \sqrt{\left(-\frac{7}{5}\right)^{2} + \left(-\frac{1}{5}\right)^{2}} = \sqrt{\frac{49}{25} + \frac{1}{25}} = \sqrt{\frac{50}{25}} = \sqrt{2}$

And for $\phi$ , we have:

$\tan \phi = \frac{-\frac{1}{5}}{-\frac{7}{5}} = \frac{1}{7}, \quad \phi = \arctan\left(\frac{1}{7}\right)$

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Questions answered

Obtain the complementary function

Find a particular integral

Combine the solutions

Analyze for large $x$

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