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navik [9.2K]
3 years ago
14

1.95(6.2-3x)-4.81=-18.46x

Mathematics
1 answer:
Citrus2011 [14]3 years ago
7 0

Answer:cdcdhfhhfhfhrhrhjrrjFtc h HD HD ( fxyfjtjtt

Step-by-step explanation:

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9
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12
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8
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9
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8
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6 0
3 years ago
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Solve the given initial-value problem. the de is of the form dy dx = f(ax + by + c), which is given in (5) of section 2.5. dy dx
shutvik [7]

\dfrac{\mathrm dy}{\mathrm dx}=\cos(x+y)

Let v=x+y, so that \dfrac{\mathrm dv}{\mathrm dx}-1=\dfrac{\mathrm dy}{\mathrm dx}:

\dfrac{\mathrm dv}{\mathrm dx}=\cos v+1

Now the ODE is separable, and we have

\dfrac{\mathrm dv}{1+\cos v}=\mathrm dx

Integrating both sides gives

\displaystyle\int\frac{\mathrm dv}{1+\cos v}=\int\mathrm dx

For the integral on the left, rewrite the integrand as

\dfrac1{1+\cos v}\cdot\dfrac{1-\cos v}{1-\cos v}=\dfrac{1-\cos v}{1-\cos^2v}=\csc^2v-\csc v\cot v

Then

\displaystyle\int\frac{\mathrm dv}{1+\cos v}=-\cot v+\csc v+C

and so

\csc v-\cot v=x+C

\csc(x+y)-\cot(x+y)=x+C

Given that y(0)=\dfrac\pi2, we find

\csc\left(0+\dfrac\pi2\right)-\cot\left(0+\dfrac\pi2\right)=0+C\implies C=1

so that the particular solution to this IVP is

\csc(x+y)-\cot(x+y)=x+1

5 0
3 years ago
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