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amid [387]
2 years ago
13

Problem 4: Solve the initial value problem

Mathematics
1 answer:
pishuonlain [190]2 years ago
7 0

Separate the variables:

y' = \dfrac{dy}{dx} = (y+1)(y-2) \implies \dfrac1{(y+1)(y-2)} \, dy = dx

Separate the left side into partial fractions. We want coefficients a and b such that

\dfrac1{(y+1)(y-2)} = \dfrac a{y+1} + \dfrac b{y-2}

\implies \dfrac1{(y+1)(y-2)} = \dfrac{a(y-2)+b(y+1)}{(y+1)(y-2)}

\implies 1 = a(y-2)+b(y+1)

\implies 1 = (a+b)y - 2a+b

\implies \begin{cases}a+b=0\\-2a+b=1\end{cases} \implies a = -\dfrac13 \text{ and } b = \dfrac13

So we have

\dfrac13 \left(\dfrac1{y-2} - \dfrac1{y+1}\right) \, dy = dx

Integrating both sides yields

\displaystyle \int \dfrac13 \left(\dfrac1{y-2} - \dfrac1{y+1}\right) \, dy = \int dx

\dfrac13 \left(\ln|y-2| - \ln|y+1|\right) = x + C

\dfrac13 \ln\left|\dfrac{y-2}{y+1}\right| = x + C

\ln\left|\dfrac{y-2}{y+1}\right| = 3x + C

\dfrac{y-2}{y+1} = e^{3x + C}

\dfrac{y-2}{y+1} = Ce^{3x}

With the initial condition y(0) = 1, we find

\dfrac{1-2}{1+1} = Ce^{0} \implies C = -\dfrac12

so that the particular solution is

\boxed{\dfrac{y-2}{y+1} = -\dfrac12 e^{3x}}

It's not too hard to solve explicitly for y; notice that

\dfrac{y-2}{y+1} = \dfrac{(y+1)-3}{y+1} = 1-\dfrac3{y+1}

Then

1 - \dfrac3{y+1} = -\dfrac12 e^{3x}

\dfrac3{y+1} = 1 + \dfrac12 e^{3x}

\dfrac{y+1}3 = \dfrac1{1+\frac12 e^{3x}} = \dfrac2{2+e^{3x}}

y+1 = \dfrac6{2+e^{3x}}

y = \dfrac6{2+e^{3x}} - 1

\boxed{y = \dfrac{4-e^{3x}}{2+e^{3x}}}

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Explanation:

The vertical sides, when read from left to right, can be divided to get this fraction: 9/90

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The fractions we constructed are equal to one another, as the triangles are said to be proportional.

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Lets cross multiply and solve for b

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9/90 = b/360

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90b = 3240

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<h3>b = 36</h3>

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A quick way to do this may be to notice how the jump from 9 to 90 is "times 10" so the jump from b to 360 is also "times 10". Think in reverse to divide 360 over 10 and we land on 36 as our answer. This line of thinking does not work as simple for all proportional problems.

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