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Okay so I am having some computational issues towards the end of the problems

Problem 1: $y' = (4x+y)^2$

$z = 4x + y$

$y = z - 4x$

$y' = \frac{dz}{dx} -4$

$z^2 = \frac{dz}{dx}-4$

$z^2 + 4 = \frac{dz}{dx}$

$\int \frac{dx}{x} = \int \frac{dz}{z^2+4}$

$ln \vert x \vert + c = \ln \vert z^2+4 \vert$

$ce^x = (4x+y)^2+4$

$y(x) = \sqrt{ce^x-4}-4x$

I don't know it seems like every single problem with this substitution is a nightmare computationally.

Second problem:

$x(x+y)y'+y(3x+y)=0$

$y' = \frac{-y(3x+y)}{x(x+y)}$

This is a homogeneous equation

$y' = -v \frac{(3x+xv)}{(x+xv)}$

$y' = -v \frac{(3+v)}{(1+v)}$

$v + \frac{dv}{dx} = \frac{-v^2-3v}{1+v}$

$x \frac{dv}{dx} = \frac{-v^2 -3v -v -v^2}{1+v}$

$x \frac{dv}{dx} = \frac{-2v^2-4v}{1+v}$

$\int \frac{1+v}{-2v(v-2)}dv = \int \frac{dx}{x}$

This a partial fractions, an ugly one at that:

$1+ v = \frac{A}{-2v}+\frac{B}{v-2}= A(v-2) -2Bv$

Now the only values of $v$ where I get any where is $v=2$ and $v = 0$ but I do not think these are viable because we get zeros in the fractions at this point.

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  • $\begingroup$ What is your question? Are you asking for someone to check your work or provide a hint? Have you checked over your work yourself yet? $\endgroup$ – kkc Aug 13 at 2:13
  • $\begingroup$ Well I cant seem to get to an answer on either one. What am I missing? Like is the work correct? Can I use those values of v? Can I just take the natural log of $z^2+4$? $\endgroup$ – K. Gibson Aug 13 at 5:17
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1.)

Your integration is not correct. $$ \int\frac{dz}{z^2+4}=\frac12\arctan(\frac z2)+C $$ so that $z=2\tan(2(x-C))$, $y=2\tan(2(x-C))-4x$.

2.)

You want to compute the coefficients in the partial fraction decomposition
$$ -\frac{1+v}{2v(v-2)}=\frac{A}{v}+\frac{B}{v-2} $$ To that end you multiply with the denominator to get $$ 1+v=-2A(v-2)-2Bv. $$ You need to keep these two forms of the equation separate, do not mix them.

Indeed then inserting $v=0$ and $v=2$ gives $4A=1$ and $4B=-3$.

But, carefully check again the signs in your formulas

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