# Solving a first order ordinary differential equation by transforming to an homogeneous equation

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. $$( 5x + 2y + 1) \, dx + ( 2x + y + 1) \, dy = 0$$
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.

• To convert the equation to an exact ODE, you need to multiply it with an integrating factor. – Shaz May 21 '20 at 19:41
• @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. – Bob May 21 '20 at 19:44
• It's really a complicated way to solve this DE. – Aryadeva May 21 '20 at 19:49

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}
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 $$y$$.