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Nutka1998 [239]

10 months ago

5

Will the soulution to the following equation be positive or negativeb÷-2=5

1 answer:

sergejj [24]10 months ago

3 0

the equation is:

$\frac{b}{-2}=5$

so we can multiply by -2 in bout sides of the equation to solve for b so:

$\frac{b\cdot(-2)}{-2}=5\cdot(-2)$

and we simplify so:

$b=-10$

So the solution will be negative.

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$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2\left(\frac yx\right)^2}{\left(1+\frac yx\right)^2}$

Substitute $v(x)=\dfrac{y(x)}x$ , so that $\dfrac{\mathrm dv(x)}{\mathrm dx}=\dfrac{x\frac{\mathrm dy(x)}{\mathrm dx}-y(x)}{x^2}$ . Then

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The remaining ODE is separable. Separating the variables gives

$\dfrac{(1+v)^2}{v(1+v^2)}\,\mathrm dv=-\dfrac{\mathrm dx}x$

Integrate both sides. On the left, split up the integrand into partial fractions.

$\dfrac{(1+v)^2}{v(1+v^2)}=\dfrac{v^2+2v+1}{v(v^2+1)}=\dfrac av+\dfrac{bv+c}{v^2+1}$

$\implies v^2+2v+1=a(v^2+1)+(bv+c)v$

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On the right, we have

$\displaystyle-\int\frac{\mathrm dx}x=-\ln|x|+C$

Solving for $v(x)$ explicitly is unlikely to succeed, so we leave the solution in implicit form,

$\ln|v(x)|+2\tan^{-1}v(x)=-\ln|x|+C$

and finally solve in terms of $y(x)$ by replacing $v(x)=\dfrac{y(x)}x$ :

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$\ln|y(x)|-\ln|x|+2\tan^{-1}\dfrac{y(x)}x=-\ln|x|+C$

$\boxed{\ln|y(x)|+2\tan^{-1}\dfrac{y(x)}x=C}$

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