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akornegay2

akornegay2

10.07.2019 •

Mathematics

find the general solution for the following homogeneous equations. show solutions
1) (xy+y2)dx−x2dy=0 (hint: let u=y/x, so y=ux, dy=xdu+udx))
2) 2x2ydx=(3x3+y3)dy. (hint: let v=x/y, so x=vy, dx=vdy+ydv)

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Ответ:

leonidas117

12.03.2022

Follow the hints.

1. Let $u=\dfrac yx$ , so that y=ux and $\mathrm dy=x,\mathrm du+u\,\mathrm dx$ . Substituting into

$(xy+y^2)\,\mathrm dx-x^2\,\mathrm dy=0$

gives

$(ux^2+u^2x^2)\,\mathrm dx-x^2(x\,\mathrm du+u\,\mathrm dx)=0$

$u^2x^2\,\mathrm dx-x^3\,\mathrm du=0$

and the remaining ODE is separable:

$x^3\,\mathrm du=u^2x^2\,\mathrm dx\implies\dfrac{\mathrm du}{u^2}=\dfrac{\mathrm dx}x$

Integrate both sides to get

$-\dfrac1u=\ln|x|+C$

$-\dfrac xy=\ln|x|+C$

$\boxed{y=\dfrac x{Cx-\ln|x|}}$

2. $Let [tex]v=\dfrac xy$ , so that x=vy and $\mathrm dx=v\,\mathrm dy+y\,\mathrm dv$ . Then

$2x^2y\,\mathrm dx=(3x^3+y^3)\,\mathrm dy$

becomes

$2v^2y^3(v\,\mathrm dy+y\,\mathrm dv)=(3v^3y^3+y^3)\,\mathrm dy$

$2v^3y^3\,\mathrm dy+2v^2y^4\,\mathrm dv=(3v^3y^3+y^3)\,\mathrm dy$

$2v^2y^4\,\mathrm dv=(v^3y^3+y^3)\,\mathrm dy$

which is separable as

$\dfrac{2v^2}{v^3+1}\,\mathrm dv=\dfrac{\mathrm dy}y$

Integrating both sides gives

$\dfrac23\ln|v^3+1|=\ln|y|+C$

$\ln|v^3+1|=\dfrac32\ln|y|+C$

$v^3+1=Cy^{3/2}$

$v=\sqrt[3]{Cy^{3/2}-1}$

$\dfrac xy=\sqrt[3]{Cy^{3/2}-1}$

$\boxed{x=y\sqrt[3]{Cy^{3/2}-1}}$

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Ответ:

abraham1366

29.03.2022

The answer for the following are

A) 14.34818

B) 10.78

0,0(0 оценок)

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