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Ulleksa [173]
3 years ago
9

Direction: Determine whether the given point is a solution to the given system of linear inequalities.

Mathematics
1 answer:
Ann [662]3 years ago
8 0

Answer:

The given point is a solution to the given system of inequalities.

Step-by-step explanation:

Again, we can substitute the coordinates of the given point into the system of inequalities. We know that the x-coordinate and y-coordinate of (3,2) are x=3 and y=2, respectively.

Plugging these values into the first inequality, y\leq x+3, gives us 2\leq 3+3, which simplifies to 2\leq 6. This is a true statement, so the given point satisfies the first inequality. We still need to check if it satisfies the second inequality though, because if it doesn't, it won't be a solution to the system.

Plugging the coordinates into the second inequality, y\geq -x+3, gives us 2\geq -3+3, which simplifies to 2\geq 0. This is also a true statement, so the given point satisfies the second inequality as well. Therefore, \bold(\bold3\bold,\bold2\bold) is a solution to the given system of inequalities since it satisfies all of the inequalities in the system. Hope this helps!

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Substitute z=\ln x, so that

\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dy}{\mathrm dz}\cdot\dfrac{\mathrm dz}{\mathrm dx}=\dfrac1x\dfrac{\mathrm dy}{\mathrm dz}

\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1x\dfrac{\mathrm dy}{\mathrm dz}\right]=-\dfrac1{x^2}\dfrac{\mathrm dy}{\mathrm dz}+\dfrac1x\left(\dfrac1x\dfrac{\mathrm d^2y}{\mathrm dz^2}\right)=\dfrac1{x^2}\left(\dfrac{\mathrm d^2y}{\mathrm dz^2}-\dfrac{\mathrm dy}{\mathrm dz}\right)

Then the ODE becomes


x^2\dfrac{\mathrm d^2y}{\mathrm dx^2}+x\dfrac{\mathrm dy}{\mathrm dx}+y=0\implies\left(\dfrac{\mathrm d^2y}{\mathrm dz^2}-\dfrac{\mathrm dy}{\mathrm dz}\right)+\dfrac{\mathrm dy}{\mathrm dz}+y=0
\implies\dfrac{\mathrm d^2y}{\mathrm dz^2}+y=0

which has the characteristic equation r^2+1=0 with roots at r=\pm i. This means the characteristic solution for y(z) is

y_C(z)=C_1\cos z+C_2\sin z

and in terms of y(x), this is

y_C(x)=C_1\cos(\ln x)+C_2\sin(\ln x)

From the given initial conditions, we find

y(1)=1\implies 1=C_1\cos0+C_2\sin0\implies C_1=1
y'(1)=8\implies 8=-C_1\dfrac{\sin0}1+C_2\dfrac{\cos0}1\implies C_2=8

so the particular solution to the IVP is

y(x)=\cos(\ln x)+8\sin(\ln x)
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