# What is the general solution of the differential equation?  (4x+3y+15)dx + (2x + y +7)dy = 0

Dec 23, 2017

$\frac{17}{34} \ln \left\mid {\left(\frac{y + 1}{x + 3}\right)}^{2} - \frac{y + 1}{x + 3} - 4 \right\mid - 5 \frac{\sqrt{17}}{34} \ln \left\mid 2 \frac{y + 1}{x + 3} + \sqrt{17} - 1 \right\mid + 5 \frac{\sqrt{17}}{34} \ln \left\mid 2 \frac{y + 1}{x + 3} - \sqrt{17} - 1 \right\mid = \ln \left\mid u \right\mid + C$

#### Explanation:

We have:

$\left(4 x + 3 y + 15\right) \mathrm{dx} + \left(2 x + y + 7\right) \mathrm{dy} = 0$

Which we can write as:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{4 x + 3 y + 15}{2 x + y + 7}$ ..... [A]

Our standard toolkit for DE's cannot be used. However we can perform a transformation to remove the constants from the linear numerator and denominator.

Consider the simultaneous equations

$\left\{\begin{matrix}4 x + 3 y + 15 = 0 \\ 2 x + y + 7 = 0\end{matrix}\right. \implies \left\{\begin{matrix}x = - 3 \\ y = - 1\end{matrix}\right.$

As a result we perform two linear transformations:

Let $\left\{\begin{matrix}u = x + 3 \\ v = y + 1\end{matrix}\right. \implies \left\{\begin{matrix}x = u - 3 \\ y = v - 1\end{matrix}\right.$

And if we substitute into the DE [A] we get

$\frac{\mathrm{dv}}{\mathrm{du}} = - \frac{4 \left(u - 3\right) + 3 \left(v - 1\right) + 15}{2 \left(u - 3\right) + \left(v - 1\right) + 7}$
$\setminus \setminus \setminus \setminus \setminus \setminus = - \frac{4 u - 12 + 3 v - 3 + 15}{2 u - 6 + v - 1 + 7}$
$\setminus \setminus \setminus \setminus \setminus \setminus = - \frac{4 u + 3 v}{2 u + v}$ ..... [B]

This is now in a form that we can handle using a substitution of the form $v = w u$ which if we differentiate wrt $u$ using the product gives us:

$\frac{\mathrm{dv}}{\mathrm{du}} = \left(w\right) \left(\frac{d}{\mathrm{du}} u\right) + \left(\frac{d}{\mathrm{du}} w\right) \left(u\right) = w + u \frac{\mathrm{dw}}{\mathrm{du}}$

Using this substitution into our modified DE [B] we get:

$\setminus \setminus \setminus \setminus \setminus w + u \frac{\mathrm{dw}}{\mathrm{du}} = - \frac{4 u + 3 w u}{2 u + w u}$

$\therefore w + u \frac{\mathrm{dw}}{\mathrm{du}} = - \frac{4 + 3 w}{2 + w}$

$\therefore u \frac{\mathrm{dw}}{\mathrm{du}} = - \frac{4 + 3 w}{2 + w} - w$

$\therefore u \frac{\mathrm{dw}}{\mathrm{du}} = - \frac{\left(4 + 3 w\right) - w \left(2 + w\right)}{2 + w}$

$\therefore u \frac{\mathrm{dw}}{\mathrm{du}} = \frac{{w}^{2} - w - 4}{2 + w}$

This is now a separable DE, so we can rearrange and separate the variables to get:

$\int \setminus \frac{2 + w}{{w}^{2} - w - 4} \setminus \mathrm{dw} = \int \setminus \frac{1}{u} \setminus \mathrm{du}$

The LHS is a standard integral, and the RHGS integral can be split into partial fractions (omitted) and then it is readily integrable, and we find that:

$\frac{17}{34} \ln \left\mid {w}^{2} - w - 4 \right\mid - 5 \frac{\sqrt{17}}{34} \ln \left\mid 2 w + \sqrt{17} - 1 \right\mid + 5 \frac{\sqrt{17}}{34} \ln \left\mid 2 w - \sqrt{17} - 1 \right\mid = \ln \left\mid u \right\mid + C$

Then restoring the earlier substitution we have:

$w = \frac{v}{u} = \frac{y + 1}{x + 3}$

Thus:

$\frac{17}{34} \ln \left\mid {\left(\frac{y + 1}{x + 3}\right)}^{2} - \frac{y + 1}{x + 3} - 4 \right\mid - 5 \frac{\sqrt{17}}{34} \ln \left\mid 2 \frac{y + 1}{x + 3} + \sqrt{17} - 1 \right\mid + 5 \frac{\sqrt{17}}{34} \ln \left\mid 2 \frac{y + 1}{x + 3} - \sqrt{17} - 1 \right\mid = \ln \left\mid u \right\mid + C$

This can be further simplified if desired.