Question a1a8c

Nov 27, 2016

${\Pi}_{k = 1}^{3} {\left(\frac{y}{x} + {r}_{k}\right)}^{{a}_{k}} = {C}_{1} / x$

Explanation:

Making $y \left(x\right) = \lambda \left(x\right) x$ and substituting into the differential equation we obtain

$\frac{\mathrm{dl} a m b \mathrm{da}}{\mathrm{dx}} = - \frac{{\lambda}^{3} + 6 \lambda + 5}{x \lambda \left(\lambda + 5\right)}$. This differential equation is separable so

$\frac{\left({\lambda}^{2} + 5 \lambda\right) \mathrm{dl} a m b \mathrm{da}}{{\lambda}^{3} + 6 \lambda + 5} = - \frac{\mathrm{dx}}{x}$

This equation can be expanded as

$\left({a}_{1} / \left(\lambda + {r}_{1}\right) + {a}_{2} / \left(\lambda + {r}_{2}\right) + {a}_{3} / \left(\lambda + {r}_{3}\right)\right) \mathrm{dl} a m b \mathrm{da} = - \frac{\mathrm{dx}}{x}$

where ${r}_{1} , {r}_{2} , {r}_{3}$ are the roots of ${\lambda}^{3} + 6 \lambda + 5 = 0$

After integrating we have

${\sum}_{k = 1}^{3} {a}_{k} \log \left\mid \lambda + {r}_{k} \right\mid = - \log \left\mid x \right\mid + C$

or

${\Pi}_{k} {\left(\lambda + {r}_{k}\right)}^{{a}_{k}} = {C}_{1} / x$

or

${\Pi}_{k = 1}^{3} {\left(\frac{y}{x} + {r}_{k}\right)}^{{a}_{k}} = {C}_{1} / x$

We should consider also the solution $y = 0$ eliminated in the substitution process.

Nov 28, 2016

See explanation

Explanation:

This is not an exact differential equation. As the coefficients of dx and dy

are homogeneous functions of x and y, the substitution y = vx would

help solving this differential equation.

Eliminating y,

$v + x \frac{\mathrm{dv}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{5 + {v}^{2}}{5 v + {v}^{2}}$

Separating variables and integrating,

$\int \frac{\mathrm{dx}}{x} = - \int \frac{1}{v + \frac{5 v + {v}^{2}}{5 + {v}^{2}}} \mathrm{dv}$

$\ln x = - \int \frac{5 + {v}^{2}}{v \left(5 v + {v}^{2}\right) + 5 + {v}^{2}} \mathrm{dv}$

$= - \int \frac{5 + {v}^{2}}{{v}^{3} + 6 {v}^{2} + 5} \mathrm{dv}$

Note that there is only one ( negative ) real zero ${v}_{1}$ for ${v}^{3} + 6 {v}^{2} + 5$

Assume that ${v}^{3} + 6 {v}^{2} + 5 = \left(v - {v}_{1}\right) \left({\left(v - \alpha\right)}^{2} + {\beta}^{2}\right)$ and

resolve the integrand into partial fractions, in the form

$\frac{P}{v - {v}_{1}} + \frac{Q \left(v - \alpha\right) + R}{{\left(v - \alpha\right)}^{2} + {\beta}^{2}}$,

where, like ${v}_{1} , \alpha \mathmr{and} \beta$,

P, Q and R are known constants. Now, the solution is

$\ln x = - \ln {\left(v - {v}_{1}\right)}^{P} - \int \frac{Q \left(v - \alpha\right) + R}{{\left(v - \alpha\right)}^{2} + {\beta}^{2}} \mathrm{dv}$.

Upon integration and rearrangement,

$\ln x {\left(v - {v}_{1}\right)}^{P} {\left({\left(v - \alpha\right)}^{2} + {\beta}^{2}\right)}^{\frac{Q}{2}} + \frac{R}{\beta} {\tan}^{- 1} \left(\frac{v - \alpha}{\beta}\right) + C$

Reverting to y,

x(y/x-v_1)^P((y/x-alpha)^2+beta^2)^(Q/2)+R/beta tan^(-1)((y/x-alpha)/beta))=C#

It is for the interested reader to evaluate the constants

${v}_{1} , \alpha , \beta , P , Q \mathmr{and} R$.

,,