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AleksAgata [21]
3 years ago
12

Use variation of parameters to find a general solution to the differential equation given that the functions y1 and y2 are linea

rly independent solutions to the corresponding homogeneous equation for t > 0.
ty'' + (2t - 1)y' - 2y = 7t2 e-2t y1 = 2t - 1, y2 = e-2t
Mathematics
1 answer:
Aleks [24]3 years ago
4 0

Recall that variation of parameters is used to solve second-order ODEs of the form

<em>y''(t)</em> + <em>p(t)</em> <em>y'(t)</em> + <em>q(t)</em> <em>y(t)</em> = <em>f(t)</em>

so the first thing you need to do is divide both sides of your equation by <em>t</em> :

<em>y''</em> + (2<em>t</em> - 1)/<em>t</em> <em>y'</em> - 2/<em>t</em> <em>y</em> = 7<em>t</em>

<em />

You're looking for a solution of the form

y=y_1u_1+y_2u_2

where

u_1(t)=\displaystyle-\int\frac{y_2(t)f(t)}{W(y_1,y_2)}\,\mathrm dt

u_2(t)=\displaystyle\int\frac{y_1(t)f(t)}{W(y_1,y_2)}\,\mathrm dt

and <em>W</em> denotes the Wronskian determinant.

Compute the Wronskian:

W(y_1,y_2) = W\left(2t-1,e^{-2t}\right) = \begin{vmatrix}2t-1&e^{-2t}\\2&-2e^{-2t}\end{vmatrix} = -4te^{-2t}

Then

u_1=\displaystyle-\int\frac{7te^{-2t}}{-4te^{-2t}}\,\mathrm dt=\frac74\int\mathrm dt = \frac74t

u_2=\displaystyle\int\frac{7t(2t-1)}{-4te^{-2t}}\,\mathrm dt=-\frac74\int(2t-1)e^{2t}\,\mathrm dt=-\frac74(t-1)e^{2t}

The general solution to the ODE is

y = C_1(2t-1) + C_2e^{-2t} + \dfrac74t(2t-1) - \dfrac74(t-1)e^{2t}e^{-2t}

which simplifies somewhat to

\boxed{y = C_1(2t-1) + C_2e^{-2t} + \dfrac74(2t^2-2t+1)}

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PLZ HELP I DONT UNDERSTAND. Can someone explain this?
DENIUS [597]

Answer:

The equation for line m is;

5y = -3x + 36

The equation of line q is;

3y = 5x + 8

Step-by-step explanation:

Looking at the diagram, firstly, we can see that while n and m are perpendicular, q and m are parallel

We should also note that the the line m and q both share the point (2,6)

Let’s try and work to get the equation of the lines

For line n, we have two points (1,-1) and (4,4)

The slope here can be obtained by the equation;

m = y2-y1/x2-x1 = (4 -(-1))/(4-1) = 5/3

Now let’s get for equation m

When two lines are perpendicular, the product of their slopes is -1

Let’s say m1 is like n slope and m2 is like m slope

m1 * m2 = -1

5/3 * m2 = -1

m2 = -3/5

So the slope of line m is -3/5

To get the equation of line m, we use the point slope form

The point to consider is (2,6)

Hence;

y-y1 = m(x-x1)

y-6 = -3/5(x-2)

5(y-6) = -3(x-2)

5y -30 = -3x + 6

5y = -3x + 6 + 30

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Since line n and q are also parallel, then their slopes are equal. Hence, the slope of line q is also 5/3

We use the point slope form to get the equation of line q

where m = 5/3 and point = (2,6)

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y-6 = 5/3(x-2)

3(y-6) = 5(x-2)

3y -18 = 5x -10

3y = 5x -10 + 18

3y = 5x + 8

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