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tangare [24]
3 years ago
13

There are x number of students at helms. If the number of students increases by 7.8% each year, how many students will be there

next year. Write an equation to express this.
Mathematics
1 answer:
vodomira [7]3 years ago
3 0

There will be 1.078x students next year and equation is number of students in next year = x + 7.8% of x

<h3><u>Solution:</u></h3>

Given, There are "x" number of students at helms.  

The number of students increases by 7.8% each year which means if there "x" number of students in present year, then the number of students in next year will be x + 7.8% of x

Number of students in next year = number of students in present year + increased number of students.

\begin{array}{l}{\text { Number of students in next year }=x+7.8 \% \text { of } x} \\\\ {\text { Number of students in next year }=x\left(1+\frac{7.8}{100}\right)} \\\\ {\text { Number of students in next year }=x(1+0.078)=1.078 x}\end{array}

Thus there will be 1.078x students in next year

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D^2(y)/(dx^2)-16*k*y=9.6e^(4x) + 30e^x
MA_775_DIABLO [31]
The solution depends on the value of k. To make things simple, assume k>0. The homogeneous part of the equation is

\dfrac{\mathrm d^2y}{\mathrm dx^2}-16ky=0

and has characteristic equation

r^2-16k=0\implies r=\pm4\sqrt k

which admits the characteristic solution y_c=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}.

For the solution to the nonhomogeneous equation, a reasonable guess for the particular solution might be y_p=ae^{4x}+be^x. Then

\dfrac{\mathrm d^2y_p}{\mathrm dx^2}=16ae^{4x}+be^x

So you have

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This means

16a(1-k)=9.6\implies a=\dfrac3{5(1-k)}
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and so the general solution would be

y=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}+\dfrac3{5(1-k)}e^{4x}+\dfrac{30}{1-16k}e^x
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