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.
Best Answer
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}