Mathematics, 27.07.2019 06:20 skylex

3. solve the differential equations a) y'' + 12y' + 32y = 0 b) y'' + 14y' + 49y = 0 c) y'' + 10y' + 34y = 0, y(0) = 1, y'(0) = 4

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Answer from: shortcake8047

$y(x)=3e^{-4x}-2e^{-8x}$

Step-by-step explanation:

I will do the first one thoroughly so you won't have any problems following to complete the rest of them.

This is a linear homogeneous second order differential, so to solve it we will use:

$y(x)=C_{1}e^{r_{1}x}+C_{2}e^{r_{2}x}$ which is a theorem that says that if r1x and r2x are both solutions off a linear homogeneous equation, and C1 and C2 are any constants, then the function above is also a solution of the equation.

We need to solve for r1 and r2 using the differential equation:

y'' + 12y' + 32y = 0

Solve the differential equation for r1 and r2 by first replacing the y'' with r^2 and the y' with r:

r^2+12r+32=0

W will factor that now to solve for the 2 values of r:

(r + 4)(r + 8) = 0

By the Zero Product Property, either one of those binomials has to equal 0 for the product to equal 0, so

r + 4 = 0 and r = -4

r + 8 = 0 and r = -8

Those are the values for r1 and r2 and we can sub them back in to the y(x) equation:

$y(x)=C_{1}e^{-4x}+C_{2}e^{-8x}$

This we will call Equation 1.

Now we find the derivative of that equation, using the rules for finding derivatives of e's:

$y'(x)=-4C_{1}e^{-4x}-8C_{2}e^{-8x}$

This we will call Equation 2.

Now we will use our first initial condition in Equation 1, where y(0) = 1:

$y(0)=C_{1}e^{(-4)(0)}+C_{2}e^{(-8)(0)}=1$

Simplifying gives you:

$y(0)=C_{1}e^0+C_{2}e^0=1$ so

$C_{1}+C_{2}=1$

Now we will use the second initial condition in Equation 2, where y'(0) = 4:

$y'(0)=-4C_{1}e^{(-4)(0)}-8C_{2}e^{(-8)(0)}=4$

Simplifying gives you:

$y'(0)=-4C_{1}e^0-8C_{2}e^0=4$ so

$-4C_{1}-8C_{2}=4$

We will now go back to the first bold equation and solve it for C1:

$C_{1}=1-C_{2}$ and sub that value in to the second bold equation to solve for C2:

$-4(1-C_{2})-8C_{2}=4$ and

$-4+4C_{2}-8C_{2}=4$ and

$-4C_{2}=8$ so

$C_{2}=-2$

Now sub that back in to the first bold equation to solve for C1:

$C_{1}-2=1$ so

$C_{1}=3$

Finally we go back to the y(x) equation and fill everything in:

$y(x)=3e^{-4x}-2e^{-8x}$

And that's your original equation! Follow this to the "t" and you'll have no problems with the other 2. They are identical in execution.

Answer from: Quest

an advantage of the standard deviation is that it increases as the dispersion of the data increases. e. the interquartile range is preferred when the data are not skewed or no have outliers. an advantage of the standard deviation is that it uses all the observations in its computation.

Answer from: Quest

Graph a should be your answer

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3. solve the differential equations a) y'' + 12y' + 32y = 0 b) y'' + 14y' + 49y = 0 c) y'' + 10y' +...

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