Write this equation as (x2+3y2)dy=xydx⇒dydx=xyx2+3y2=yx1+3y2x2 (In the last equality, we divided the numerator and denominator by x2 .) Therefore, the equation becomes is dydx=yx1+3y2x2 (1) which is of the form dydx=h(yx) Thus, we will use the substitution y=ux⇒dydx=dudx⋅x+u (2) Plugging (2) into (1), we get that, dudx⋅x+u=u1+3u2⇒xdudx=u1+3u2−u=u−u(1+3u2)1+3u2=−3u21+3u2 Therefore, xdudx=−3u21+3u2⇒1+3u23u3du=−dxx Integrating both sides: ∫1+3u23u3du=∫−dxx Now, ∫−dxx=−ln|x|+C2=ln1|x|+C2 where C2 is some constant. (We have also used that n⋅ln⋅y=ln⋅yn) On the other hand, ∫1+3u23u3du=13∫duu3+∫duu =13∫u−3du+ln|u| =−16u−2+ln|u|+C1 =−16u2+ln|u|+C1 Here C1 is some constant Using (3) we now get that −16u2+ln|u|+C1=ln1|x|+C2 Defining C=C2−C1, we get −16u2+ln|u|=ln1|x|+C Finally, recall that u=yx to write the final solution (in the implicit form) ln|yx|−x26y2=ln1|x|+C Answer ln|yx|−x26y2=ln1|x|+C

Key to " xydx−(x2+3y2)dy=0"

avissidep

Answered question

2020-10-26

$x y d x - (x^{2} + 3 y^{2}) d y = 0$

Answer & Explanation

yunitsiL

Skilled2020-10-27Added 108 answers

Write this equation as

(x^{2} + 3 y^{2}) d y = x y d x \Rightarrow \frac{d y}{d x} = \frac{x y}{x^{2} + 3 y^{2}} = \frac{\frac{y}{x}}{1 + 3 \frac{y^{2}}{x^{2}}}

(In the last equality, we divided the numerator and denominator by

x^{2}

.) Therefore, the equation becomes is

\frac{d y}{d x} = \frac{\frac{y}{x}}{1 + 3 \frac{y^{2}}{x^{2}}}

(1) which is of the form

\frac{d y}{d x} = h (\frac{y}{x})

Thus, we will use the substitution

y = u x \Rightarrow \frac{d y}{d x} = \frac{d u}{d x} \cdot x + u

(2)
Plugging (2) into (1), we get that,

\frac{d u}{d x} \cdot x + u = \frac{u}{1 + 3 u^{2}} \Rightarrow x \frac{d u}{d x} = \frac{u}{1 + 3 u^{2}} - u = \frac{u - u (1 + 3 u^{2})}{1 + 3 u^{2}} = - \frac{3 u^{2}}{1 + 3 u^{2}}

Therefore,

x \frac{d u}{d x} = - \frac{3 u^{2}}{1 + 3 u^{2}} \Rightarrow \frac{1 + 3 u^{2}}{3 u^{3}} d u = - \frac{d x}{x}

Integrating both sides:

\int \frac{1 + 3 u^{2}}{3 u^{3}} d u = \int - \frac{d x}{x}

Now,

\int - \frac{d x}{x} = - l n | x | + C_{2} = l n \frac{1}{| x |} + C_{2}

where

C_{2}

is some constant. (We have also used that

n \cdot l n \cdot y = l n \cdot y^{n}

)
On the other hand,

\int \frac{1 + 3 u^{2}}{3 u^{3}} d u = \frac{1}{3} \int \frac{d u}{u^{3}} + \int \frac{d u}{u}

= \frac{1}{3} \int u^{- 3} d u + l n | u |

= - \frac{1}{6} u^{- 2} + l n | u | + C_{1}

= - \frac{1}{6 u^{2}} + l n | u | + C_{1}

Here

C_{1}

is some constant
Using (3) we now get that

- \frac{1}{6 u^{2}} + l n | u | + C_{1} = l n \frac{1}{| x |} + C_{2}

Defining

C = C_{2} - C_{1}

, we get

- \frac{1}{6 u^{2}} + l n | u | = l n \frac{1}{| x |} + C

Finally, recall that

u = \frac{y}{x}

to write the final solution (in the implicit form)

l n | \frac{y}{x} | - \frac{x^{2}}{6 y^{2}} = l n \frac{1}{| x |} + C

Answer

l n | \frac{y}{x} | - \frac{x^{2}}{6 y^{2}} = l n \frac{1}{| x |} + C

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