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2 years ago

5

Which line will have no solution with the parabola y – x + 2 = x2? y = –3x –3 y = –2x –3 y = 2x –3 y = 3x –3

1 answer:

jeka942 years ago

3 0

Try this option:
the line y=-2x-3.
Explanation:
if to see the pair of equations: y=x²+x-2 and y=-3x-3, ⇔ x²+x-2=-3x-3, ⇔ x²+4x+1=0, D>0, then they have two common points.
if to see the pair of equations: y=x²+x-2 and y=-2x-3, ⇔ x²+3x+1=0, ⇒D>0, then they have two common points.
for y=x²+x-2 and y=2x-3: x²+x+1=0, D<0, it means, that these equations have no common points.
for y=x²+x-2 and y=3x-3: x²-2x+1=0, D=0, it means, that these equations have the only common point.

answer: y=2x-3.

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Answer:

$y(x)=sin(2x)+2cos(2x)$

Step-by-step explanation:

$y''+4y=0$

This is a homogeneous linear equation. So, assume a solution will be proportional to:

$e^{\lambda x} \\\\for\hspace{3}some\hspace{3}constant\hspace{3}\lambda$

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Using the characteristic equation:

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Factor out $e^{\lambda x}$

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Where:

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Therefore the zeros must come from the polynomial:

$\lambda^2+4 =0$

Solving for $\lambda$ :

$\lambda =\pm2i$

These roots give the next solutions:

$y_1(x)=c_1 e^{2ix} \\\\and\\\\y_2(x)=c_2 e^{-2ix}$

Where $c_1$ and $c_2$ are arbitrary constants. Now, the general solution is the sum of the previous solutions:

$y(x)=c_1 e^{2ix} +c_2 e^{-2ix}$

Using Euler's identity:

$e^{\alpha +i\beta} =e^{\alpha} cos(\beta)+ie^{\alpha} sin(\beta)$

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Redefine:

$i(c_1-c_2)=c_1\\\\c_1+c_2=c_2$

Since these are arbitrary constants

$y(x)=c_1sin(2x)+c_2cos(2x)$

Now, let's find its derivative in order to find $c_1$ and $c_2$

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Evaluating $y(0)=2$ :

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Finally, the solution is given by:

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Step-by-step explanation:

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