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The following problem is from the book "Introduction to Ordinary Differential Equations" By Shepley L. Ross. It is problem number 7 in section 2.4.

Problem:
Solve the following differential equations by making a suitable transformation. $$ ( 5x + 2y + 1) \, dx + ( 2x + y + 1) \, dy = 0 $$
Answer:
We want to transform this to a homogeneous differential equation. So we set up the following system of linear equations. \begin{align*} 5x + 2y + 1 &= 0 \\ 2x + y + 1 &= 0 \\ y &= -2x - 1 \\ 5x + 2 ( -2x - 1 ) + 1 &= 0 \\ 5x - 4x - 2 + 1 &= 0 \\ x &= 1 \\ y &= -2(1) - 1 = -3 \\ x_1 &= x - 1 \\ y_1 &= y + 3 \\ ( 5(x_1 + 1) + 2(y_1 - 3) + 1) \, dx + ( 2(x_1 + 1) + y_1 - 3 + 1) \, dy = 0 \\ ( 5x_1 + 5 + 2y_1 - 6 + 1) \, dx + ( 2x_1 + 2 + y_1 - 3 + 1) \, dy = 0 \\ \end{align*} Now we note that $dx_1 = dx$ and $dy_1 = dy$. \begin{align*} ( 5x_1 + 2y_1 ) \, dx_1 + ( 2x_1 + y_1 ) \, dy_1 &= 0 \\ ( 5x_1 + 2y_1 ) \, dx_1 &= -( 2x_1 + y_1 ) \, dy_1 \\ \frac{dy_1}{dx_1} &= -\frac{ 5x_1 + 2y_1 }{2x_1 + y_1 } = -\frac{ 5 + 2\left( \frac{y_1}{x_1}\right) }{2 + \left( \frac{y_1}{x_1} \right) } \end{align*} Now let $v = \frac{y_1}{x_1}$. This gives us: \begin{align*} y_1 &= x_1 v \\ \frac{dy_1}{dx_1}&= x_1 \frac{dv}{dx_1} + v \\ x_1 \frac{dv}{dx_1} + v &= - \frac{5 + 2v}{2 + v} \\ x_1 \frac{dv}{dx_1} &= \frac{-5 - 2v - 2 - v}{2 + v} = - \frac{3v+7}{v+2} \\ \frac{v+2}{3v+7} \, dv &= - \frac{dx_1}{x} \end{align*} Now we need to integrate both sides. Using an online integral calculator, I find: $$ \int \frac{v+2}{3v+7} \, dv = \frac{ - \ln{|3v+7|} }{9} - \frac{v}{3} - C_1 $$ \begin{align*} \frac{ - \ln{|3v+7|} }{9} - \frac{v}{3} - C_1 &= -\ln{|x_1|} \\ \ln{|3v+7|} + 3v + C_1 &= 9\ln{|x_1|} \\ \ln{|3v+7|} + \ln{e^{3v}} + C_1 &= 9\ln{|x_1|} \\ \ln{|3v+7|} + \ln{e^{3v}} - 9\ln{|x_1|} &= -C_1 \\ \frac{(3v+7)e^{3v}}{x_1^9} &= C_2 \end{align*} The book's answer is: $$ 5x^2 + 4xy + y^2 + 2x + 2y = c $$ My answer is not going to match. Where did I go wrong?

Here is my second attempt to solve the problem. I believe I have it right now. If I do not, please tell me I am still wrong.

We want to transform this to a homogeneous differential equation. So we set up the following system of linear equations. \begin{align*} 5x + 2y + 1 &= 0 \\ 2x + y + 1 &= 0 \\ y &= -2x - 1 \\ 5x + 2 ( -2x - 1 ) + 1 &= 0 \\ 5x - 4x - 2 + 1 &= 0 \\ x &= 1 \\ y &= -2(1) - 1 = -3 \\ x_1 &= x - 1 \\ y_1 &= y + 3 \\ ( 5(x_1 + 1) + 2(y_1 - 3) + 1) \, dx + ( 2(x_1 + 1) + y_1 - 3 + 1) \, dy = 0 \\ ( 5x_1 + 5 + 2y_1 - 6 + 1) \, dx + ( 2x_1 + 2 + y_1 - 3 + 1) \, dy = 0 \\ \end{align*} Now we note that $dx_1 = dx$ and $dy_1 = dy$. \begin{align*} ( 5x_1 + 2y_1 ) \, dx_1 + ( 2x_1 + y_1 ) \, dy_1 &= 0 \\ ( 5x_1 + 2y_1 ) \, dx_1 &= -( 2x_1 + y_1 ) \, dy_1 \\ \frac{dy_1}{dx_1} &= -\frac{ 5x_1 + 2y_1 }{2x_1 + y_1 } = -\frac{ 5 + 2\left( \frac{y_1}{x_1}\right) }{2 + \left( \frac{y_1}{x_1} \right) } \end{align*} Now let $v = \frac{y_1}{x_1}$. This gives us: \begin{align*} y_1 &= x_1 v \\ \frac{dy_1}{dx_1}&= x_1 \frac{dv}{dx_1} + v \\ x_1 \frac{dv}{dx_1} + v &= - \frac{5 + 2v}{2 + v} \\ x_1 \frac{dv}{dx_1} &= \frac{-5 -2v - v(v+2)}{v+2} = \frac{-5 -2v - v^2 - 2v}{v+2} \\ -x_1 \frac{dv}{dx_1} &= \frac{v^2+4v+5}{v+2} \\ \frac{ (v+2) \, \, dv }{v^2 + 4v + 5} &= -\frac{dx_1}{x} \end{align*} Using an online integral calculator, I find: $$ \int \frac{v+2}{v^2+4v+5} \,\, dv = \frac{ \ln{(v^2 + 4v + 5)}}{2} + C_1$$ \begin{align*} - \ln{|x_1|} &= \frac{ \ln{(v^2 + 4v + 5)}}{2} + C_1 \\ - 2 \ln{|x_1|} &= \ln{( v^2 + 4v + 5)} + 2C_1 \\ \ln{(v^2 + 4v + 5)} + 2 \ln{|x_1|} &= -2C_1 \\ x_1 ^2 ( v^2 + 4v + 5) &= C_2 \\ x_1^2 \left( \frac{y_1^2}{x_1^2} + \frac{4y_1}{x_1} + 5 \right) &= C_2 \\ y_1^2 + 4x_1 y_1 + 5x_1 ^2 &= C_2 \\ (y+3)^2 + 4(x-1)(y+3) + 5(x-1)^2 &= C_2 \\ y^2 + 6y + 9 + 4(xy + 3x - y - 3) + 5(x^2 - 2x + 1) &= C_2 \\ 5x^2 + 4xy + 2x + 2y + y^2 + 2 &= C_2 \\ \end{align*} Hence the answer we seek is: $$ 5x^2 + 4xy + y^2 + 2x + 2y = c $$ The book's answer is: $$ 5x^2 + 4xy + y^2 + 2x + 2y = c $$ My answer now matches.

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    $\begingroup$ To convert the equation to an exact ODE, you need to multiply it with an integrating factor. $\endgroup$ – Shaz May 21 '20 at 19:41
  • $\begingroup$ @Shaz The title of the post was bad, I fixed it. I believe my general approach to solving the differential equation is correct. If not, please tell me. $\endgroup$ – Bob May 21 '20 at 19:44
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    $\begingroup$ It's really a complicated way to solve this DE. $\endgroup$ – Aryadeva May 21 '20 at 19:49
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You're mistake is towards the end (fortunately): \begin{align} x_1\dfrac{dv}{dx_1}+v & =-\dfrac{5+2v}{2+v}\\ -x_1\dfrac{dv}{dx_1}-v & =\dfrac{5+2v}{2+v}\\ -x_1\dfrac{dv}{dx_1}& =\dfrac{5+2v+v(2+v)}{2+v}\\ -x_1\dfrac{dv}{dx_1} & =\dfrac{5+4v+v^2}{2+v}\\ \end{align}

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An error is $$ -\frac{5+2v}{2+v} - v = \frac{-5-4v-v^2}{2+v} \text{,} $$ not $\cdots = \frac{-5-2v-2-v}{2+v}$.

The resulting integrals are then logarithms which have some hope of simplifying to a polynomial in $x$ and $y$.

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