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3 years ago

3x +y = 14 x-5y = 15

Mathematics

1 answer:

erik [133]3 years ago

3 0

Answer:

y=-3x+14, y= x/5 -3

Step-by-step explanation:

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From the homogeneous part of the ODE, we can get two fundamental solutions. The characteristic equation is

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$y_1=e^x\cos x$
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The particular solution is easiest to obtain via variation of parameters. We're looking for a solution of the form

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where

$u_1=-\displaystyle\frac13\int\frac{y_2e^x\sec x}{W(y_1,y_2)}\,\mathrm dx$
$u_2=\displaystyle\frac13\int\frac{y_1e^x\sec x}{W(y_1,y_2)}\,\mathrm dx$

and $W(y_1,y_2)$ is the Wronskian of the fundamental solutions. We have

$W(e^x\cos x,e^x\sin x)=\begin{vmatrix}e^x\cos x&e^x\sin x\\e^x(\cos x-\sin x)&e^x(\cos x+\sin x)\end{vmatrix}=e^{2x}$

and so

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Therefore the particular solution is

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so that the general solution to the ODE is

$y=C_1e^x\cos x+C_2e^x\sin x+\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x$

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