Question

In the following, show that the given differential equation is homogeneous and solve each of them.
1. $(x^2+xy)dy=(x^2+y^2)dx$
2. $y^{\prime }=\frac{x+y}{x}$
3. $\left(xy\right)dy(x+y)dx=0$
4. $(x^2+y^2)dx+2xydy=0$
5. $x^2\frac{dy}{dx}=x^2-2y^2+xy$
6. $xdy.ydx=\sqrt{x^2+y^2}dx$
7. $\{x\cos \left(\frac{y}{x}\right)+y\sin \left(\frac{y}{x}\right)\}ydx=\{y\sin \left(\frac{y}{x}\right)-x\cos \left(\frac{y}{x}\right)\}xdy$
8. $x\frac{dy}{dx}-y+x\sin \left(\frac{y}{x}\right)=0$
9. $ydx+x\log \left(\frac{y}{x}\right)dy-2xdy=0$

Answer 1

1) $(x^2+xy)dy–(x^2+y^2)dx$

$\frac{dy}{dx}=\frac{x^2+y^2}{x^2+xy}=\frac{1+(\frac{y}{x}{)}^2}{1+\left(\frac{y}{x}\right)}$

$u=\frac{y}{x}$

$\frac{dy}{dx}=\frac{1}{x}\left(\frac{dy}{dx}\right)-\frac{y}{x^2}=\frac{1}{x}\frac{dy}{dx}-\left(\frac{u}{x}\right)$

$(\frac{du}{dx}+\frac{u}{x})x=\frac{dy}{dx}$

$(\frac{xdu}{dx}+4)=\frac{1+4^2}{1+4}$

$\frac{xdu}{dx}=\frac{1+4^2}{1+4}-u=\frac{1+4^2-4-4^2}{1+4}$

$\frac{xdu}{dx}=\left(\frac{1-4}{1+4}\right)$

$\Rightarrow \left(\frac{1+4}{1-4}\right)du=\frac{dx}{x}$

$\int \frac{-4-1}{1-4}dx=-\int \frac{dx}{x}$

$=\int \frac{-4+1-1-1}{1-4}du=-\int \frac{dx}{x}$

$=\int (1-\frac{2}{1-4})du=-\int \frac{dx}{x}$

$u+2ln(1-y)=-ln\left(x\right)+c$

$\Rightarrow u+2ln(1-4)=-ln\left(x\right)+c$

$\Rightarrow u=\frac{y}{x}$

$\Rightarrow \frac{y}{x}=2ln(1-\frac{y}{x})=-ln\left(x\right)+c$

$\Rightarrow y+2xln(1-\frac{y}{x})=-xln\left(x\right)+xc$

2) $y^1=\frac{x+y}{x}$

$y^1=1+\frac{y}{x}$

$\frac{y}{x}=u\Rightarrow \left(\frac{dy}{dx}\right)=4+x\frac{du}{dx}$

$\frac{xdu}{dx}=1\Rightarrow \int du=\int \frac{dx}{x}$

$u=ln\left(x\right)+c$

$\frac{y}{x}=ln\left(x\right)+c\Rightarrow y=xln\left(x\right)+cx$

3) wrong equation

4) $(x^2+y^2)dx+2xydy=0$

$\frac{dy}{dx}=\frac{-(x^2+y^2)}{2xy}$

$\frac{dy}{dx}=\frac{-(1+{\left(\frac{y}{x}\right)}^2)}{2\left(\frac{y}{x}\right)}$

$\frac{y}{x}=u\Rightarrow \frac{dy}{dx}=u=x\frac{du}{dx}$

$u=\frac{xdu}{dx}=\frac{-(1+4^2}{24}$

$\frac{xdu}{dx}=\frac{-(1+4^2)}{24}–4=\frac{-1-4^2-{24}^2}{24}$

$\frac{xdu}{dx}=\frac{-1-{34}^2}{24}$

$\int \frac{24}{1+{34}^2}du=-\int \frac{dx}{x}$

$\int \frac{dt}{1+3t}=-\int \frac{dx}{x}$

$\frac{1}{3}ln(1+3t)=-ln\left(x\right)+ln\left(c\right)$

$\frac{1}{3}ln(1+3t)=+ln\left(\frac{C}{x}\right)$

$\frac{1}{3}ln(1+3x^2)=ln\left(\frac{c}{x}\right)$

$(1+3x^2{)}^{1/3}=\left(\frac{C}{x}\right)$

5) $x^2\frac{dy}{dx}=x^2-2y^2+xy$

$\frac{dy}{dx}=1-2{\left(\frac{y}{x}\right)}^2+\left(\frac{y}{x}\right)$

$\frac{y}{x}=4\Rightarrow \frac{dy}{dx}=4+x\frac{dy}{dx}$

$4+x\frac{dy}{dx}=1-2y^2+4$

$x\frac{dy}{dx}=1-{24}^2+0$

$\int \frac{dy}{1-{24}^2}=\int \frac{dx}{x}$

$\int \frac{du}{1-(\sqrt{24}{)}^2}=\int \frac{dx}{x}$

$\sqrt{24}=+7\sqrt{2}du=d+$

$\Rightarrow \frac{1}{\sqrt{2}}\int \frac{d+}{1-+2}=\int \frac{dx}{x}$

$=\frac{1}{\sqrt{2}}[\int (\frac{1/2}{1-+}+\frac{\left(1/2\right)}{1+1})dt]=ln\left(x\right)+c$

$\frac{1}{12}\left[\frac{-1}{2}ln\right(1-+)+\frac{1}{2}ln(1+1\left)\right]=ln\left(x\right)+c$

$\frac{1}{12}\left[\frac{-1}{2}ln\right(1-4^2)+\frac{1}{2}ln(1+4^2\left)\right]=ln\left(x\right)+c$

$\Rightarrow \frac{1}{12}\times \frac{1}{2}ln\left(\frac{1+4^2}{1-4^2}\right)-ln+c$

$\frac{1}{2\sqrt{x}}ln\left(\frac{x^2+y^2}{x^2-y^2}\right)=lnx+c$

6) wrong question

7) $[x\cos \left(\frac{y}{x}\right)+y\sin \left(\frac{y}{x}\right)]ydx=[y\sin \left(\frac{y}{x}\right)-x\cos \left(\frac{y}{x}\right)]xdy$

$[\cos \left(\frac{y}{x}\right)+\left(\frac{y}{x}\right)\sin \left(\frac{y}{x}\right)]\left(\frac{y}{x}\right)dx=[\frac{y}{x}\sin \left(\frac{y}{x}\right)-\cos \left(\frac{y}{x}\right)]dy$

$\frac{dy}{dx}=\left(\frac{y}{x}\right)\left[\frac{\cos \left(\frac{y}{x}\right)+\left(\frac{y}{x}\right)\sin \left(\frac{y}{x}\right)}{-\cos \left(\frac{y}{x}\right)+\left(\frac{y}{x}\right)\sin \left(\frac{y}{x}\right)}\right]$

$\frac{y}{x}=u=\frac{dy}{dx}=u+x\frac{du}{dx}$
$y+x\frac{dy}{dx}=4\left[\frac{\cos u+u\sin u}{-\cos u+u\sin u}\right]$

$s\frac{dy}{dx}=4[\frac{\cos u+4\sin u}{u\sin u-\cos u}-1]$

$=4\left[\frac{\cos u+u\sin u-u\sin u+\cos 4}{u\sin u-\cos u}\right]$

$x\frac{dy}{dx}=\frac{2u\cos u}{u\sin u-\cos u}$

$\left(\frac{u\sin u-\cos u}{u\cos u}\right)du=2dx$

$\int (\tan u-\frac{1}{4})du=\int 2dx$

$\Rightarrow ln|\sec u|-ln\left(u\right)=2x+c$

$\Rightarrow ln\left(\frac{\sec u}{u}\right)=2x+c$

$\Rightarrow \left(\frac{\sec \left(\frac{y}{x}\right)}{\left(\frac{y}{x}\right)}\right)=2x+c$

$\Rightarrow ln\left(\frac{x\sec \left(\frac{y}{x}\right)}{y}\right)=2x+c$

8) $x\frac{dy}{dx}-y+x\sin \left(\frac{y}{x}\right)=0$

$\frac{dy}{dx}-\frac{y}{x}+\sin \left(\frac{y}{x}\right)=0$

$u=\frac{y}{x}\Rightarrow \frac{dy}{dx}=u+x\frac{du}{dx}$

$\frac{dy}{dx}=\frac{y}{x}-\sin \left(\frac{y}{x}\right)$

$u+x\frac{dy}{dx}=u-\sin \left(u\right)$

$\Rightarrow x\frac{dy}{dx}=-\sin u\Rightarrow \int \frac{dy}{dx}=-\int \frac{dx}{x}$

$\int cosecudu=-\int \frac{dx}{x}$

$\Rightarrow \log \int cosecu-\cot u=-ln\left(x\right)+ln\left(c\right)$

$\log ln|\cos u-\cot u|=ln|\frac{x}{c}|$

$(cosecu-\cot u).x=c$

$x\left[cose\right(\frac{y}{x})-\cot (\frac{y}{x}\left)\right]=c$

9) $ydx+x\log \left(\frac{y}{x}\right)dy-2xdy=0$

$(\frac{y}{x}dx+\left[\log \right(\frac{y}{x}-2]dy=0$

$\frac{dy}{dx}=\frac{(\frac{y}{x}}{2-\log \left(\frac{y}{x}\right)}$

$\frac{y}{x}=u\Rightarrow \frac{dy}{dx}=u+x\frac{du}{dx}$

$u+x\frac{dy}{dx}=\frac{u}{2-\log \left(u\right)}$

$x\frac{dy}{dx}=\frac{4-24+4\log u}{2-\log u}$

$=\frac{-4+u\log u}{2-\log u}$

$\frac{2-\log u}{4\log u-4}du=\frac{dx}{x}$

$\frac{2-\log u}{u(\log u-1)}du=\frac{dx}{x}$

$\log u=+$

$\frac{1}{4}du=dt$

$\frac{2-1}{4-1}dt=\frac{dx}{x}$

$\left(\frac{+-1-1}{4-1}\right)dt=\frac{-dx}{x}$

$\int \left(\frac{1}{1-1}\right)dt=\int -\frac{dx}{x}$

$4-ln(4-1)=-ln\left(x\right)+c$

$\log u=\log (\log u-1)=-ln\left(x\right)+c$

$\log \left(\frac{y}{x}\right)-\log \left(\log \right(\frac{y}{x}-1)=-ln(x)+c$