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#### karush

##### Well-known member

- Jan 31, 2012

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\(\displaystyle \frac{dy}{dx}=1+\frac{y}{x}+\left(\frac{y}{x}\right)^2\)

Next consider the substitution:

\(\displaystyle v=\frac{y}{x}\implies y=vx\implies \frac{dy}{dx}=x\frac{dv}{dx}+v\)

So, make the substitutions and simplify...what do you get?

- Jan 29, 2012

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(Notice that both numerator and denominator consist of terms that are quadratic in x and y. That's why we can cancel \(\displaystyle t^2\).)

And the point of that is that we can replace y with y/x and get a simpler equation. Dividing, like MarkFL did, both numerator and denominator on the right by \(\displaystyle x^2\) we get \(\displaystyle \frac{dy}{dx}= \frac{1+ \frac{y}{x}+ \left(\frac{y}{x}\right)^2}{1}= \left(\frac{y}{x}\right)^2+ \frac{y}{x}+ 1\).

Letting \(\displaystyle u= \frac{y}{x}\), \(\displaystyle y= xu\) so \(\displaystyle \frac{dy}{dx}= x\frac{du}{dx}+ u\).

The equation becomes \(\displaystyle x\frac{du}{dx}+ u= u^2+ u+ 1\).

\(\displaystyle x\frac{du}{dx}= u^2+ 1\), a "separable equation".

We can write that as \(\displaystyle \frac{du}{u^2+ 1}= \frac{dx}{x}\) and integrate both sides.

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- Jan 31, 2012

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the LHS looks like a product... I think!

\(\displaystyle \frac{dy}{dx}=1+\frac{y}{x}+\left(\frac{y}{x}\right)^2\)

Next consider the substitution:

\(\displaystyle v=\frac{y}{x}\implies y=vx\implies \frac{dy}{dx}=x\frac{dv}{dx}+v\)

So, make the substitutions and simplify...what do you get?

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- #5

- Jan 31, 2012

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$$\int \dfrac{1}{u^2 +1} du=\int \dfrac{1}{x}dx $$We can write that as \(\displaystyle \frac{du}{u^2+ 1}= \frac{dx}{x}\) and integrate both sides.

$$\arctan u = \ln x +c $$

replace u

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- Jan 31, 2012

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$$\int \dfrac{1}{u^2 +1} du=\int \dfrac{1}{x}dx $$

$$\arctan u = \ln x +c $$

replace $u= \frac{y}{x}$

$$\arctan u = \ln x +c $$

replace $u= \frac{y}{x}$

$\arctan \dfrac{y}{x} - \ln x =c$ I think this it thanks everyone gteat help |

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- Jan 31, 2012

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I tried plotting this with different values of c but couldn't see any difference

also how do you get "solved" in the title when your done

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Next we have:

\(\displaystyle \frac{dy}{dx}=1+\frac{y}{x}+\left(\frac{y}{x}\right)^2\)

Next consider the substitution:

\(\displaystyle v=\frac{y}{x}\implies y=vx\implies \frac{dy}{dx}=x\frac{dv}{dx}+v\)

So, make the substitutions and simplify...what do you get?

\(\displaystyle x\frac{dv}{dx}+v=1+v+v^2\)

\(\displaystyle x\frac{dv}{dx}=1+v^2\)

\(\displaystyle \int \frac{1}{v^2+1}\,\frac{dv}{dx}\,dx=\int\frac{1}{x}\,dx\)

\(\displaystyle \arctan(v)=\ln|c_1x|\)

\(\displaystyle v=\tan(\ln|c_1x|)\)

\(\displaystyle y(x)=x\tan(\ln|c_1x|)\)

- Jan 29, 2012

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Karush, do you see that your solution, $arctan(y/x)- ln(|x|)= c$, is the same as MarkFL's, $y= x tan(ln|c_1x|)$?

(Well, I added the absolute value that should have been in yours. $\int \frac{dx}{x}= ln(|x|)+ C$)

From $arctan(y/x)- ln(|x|)= c$ add ln(|x|) to both sides to get $arctan(y/x)= ln(|x|)+ c$. If we let $c_1= e^c$ (and, of course, $c_1$ is positve) the $c= ln(c_1)$ so $arctan(y/x)= ln(|x|)+ ln(c_1)= ln(|c_1x|)$.

Now, obviously, take the tangent of both sides to get $y/x= tan(ln(|c_1x|)$. Finally multiply both sides by x: $y= xtan(ln(|c_1x|)$.

(Well, I added the absolute value that should have been in yours. $\int \frac{dx}{x}= ln(|x|)+ C$)

From $arctan(y/x)- ln(|x|)= c$ add ln(|x|) to both sides to get $arctan(y/x)= ln(|x|)+ c$. If we let $c_1= e^c$ (and, of course, $c_1$ is positve) the $c= ln(c_1)$ so $arctan(y/x)= ln(|x|)+ ln(c_1)= ln(|c_1x|)$.

Now, obviously, take the tangent of both sides to get $y/x= tan(ln(|c_1x|)$. Finally multiply both sides by x: $y= xtan(ln(|c_1x|)$.

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