Find measure of angle 1

Mathematics

1 answer:

fenix001 [56]3 years ago

8 0

Just look at the top triangles
Angles in a triangle add up to 180°

To find the angle next to 36°:
It shows that all together its a 90° angle so you do 90-36= 54°

To find the angle next to 88°:
Angles on a straight line add up to 180° so you do 180-88= 92°

You now have the 2 of the angles in the triangle with angle 1

You now add these 2 angles so 54+92= 146°
and as angles in a triangle add up to 180° you do 180-146= 34°

So angle 1 is 34°

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The indicated function y1(x is a solution of the given differential equation. use reduction of order or formula (5 in section 4.

Taya2010 [7]

Given a solution $y_1(x)=\ln x$ , we can attempt to find a solution of the form $y_2(x)=v(x)y_1(x)$ . We have derivatives

$y_2=v\ln x$
${y_2}'=v'\ln x+\dfrac vx$
${y_2}''=v''\ln x+\dfrac{v'}x+\dfrac{v'x-v}{x^2}=v''\ln x+\dfrac{2v'}x-\dfrac v{x^2}$

Substituting into the ODE, we get

$v''x\ln x+2v'-\dfrac vx+v'\ln x+\dfrac vx=0$
$v''x\ln x+(2+\ln x)v'=0$

Setting $w=v'$ , we end up with the linear ODE

$w'x\ln x+(2+\ln x)w=0$

Multiplying both sides by $\ln x$ , we have

$w' x(\ln x)^2+(2\ln x+(\ln x)^2)w=0$

and noting that

$\dfrac{\mathrm d}{\mathrm dx}\left[x(\ln x)^2\right]=(\ln x)^2+\dfrac{2x\ln x}x=(\ln x)^2+2\ln x$

we can write the ODE as

$\dfrac{\mathrm d}{\mathrm dx}\left[wx(\ln x)^2\right]=0$

Integrating both sides with respect to $x$ , we get

$wx(\ln x)^2=C_1$
$w=\dfrac{C_1}{x(\ln x)^2}$

Now solve for $v$ :

$v'=\dfrac{C_1}{x(\ln x)^2}$
$v=-\dfrac{C_1}{\ln x}+C_2$

So you have

$y_2=v\ln x=-C_1+C_2\ln x$

and given that $y_1=\ln x$ , the second term in $y_2$ is already taken into account in the solution set, which means that $y_2=1$ , i.e. any constant solution is in the solution set.