# Substitution methods of first order linear eqautions [problem help]

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.

• 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? – kkc Aug 13 at 2:13
• 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 $z^2+4$? – K. Gibson Aug 13 at 5:17

Your integration is not correct. $$\int\frac{dz}{z^2+4}=\frac12\arctan(\frac z2)+C$$ so that $$z=2\tan(2(x-C))$$, $$y=2\tan(2(x-C))-4x$$.
$$-\frac{1+v}{2v(v-2)}=\frac{A}{v}+\frac{B}{v-2}$$ 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 $$4B=-3$$.