Differentiation (Part-38) | S N De

Solve : 

$~1.~~\frac{dy}{dx}=\frac{4x-5y+3}{5x+y-2}$

Sol. $~~\frac{dy}{dx}=\frac{4x-5y+3}{5x+y-2}\rightarrow(1)$

Let $~x=x'+\alpha,~~y=y'+\beta \\ \therefore \frac{dy}{dx}=\frac{dy'}{dx'}\rightarrow(2)$

Then from $~(1),(2)~$ we get,

$~~~\frac{dy'}{dx'}=\frac{4(x'+\alpha)-5(y'+\beta)+3}{5(x'+\alpha)+(y'+\beta)-2} \\ \therefore \frac{dy'}{dx'}=\frac{4x'-5y'+(4\alpha-5\beta+3)}{5x'+y'+(5\alpha+\beta-2)}\rightarrow(3)$

We choose $~~\alpha,\beta~$ such that 

$~~ 4\alpha-5\beta+3=0\rightarrow(4),\\~~ 5\alpha+\beta-2=0\rightarrow(5)$

Solving $\,(4)\,$ and $\,(5)\,$ we get,

$~~\alpha=\frac{7}{29},~~\beta=\frac{23}{29}.$

Let $~~y'=vx' \\ \Rightarrow \frac{dy'}{dx'}=v+x'\frac{dv}{dx'}\rightarrow(6)$

So, from $~~(3),(4),(5),(6)~$ we get,

$~~v+x'\frac{dv}{dx'}=\frac{4x'-5vx'}{5x'+vx'} \\ \text{or,}~~ x'\frac{dv}{dx'}=\frac{4-5v}{5+v}-v \\ \text{or,}~~ x'\frac{dv}{dx'}=\frac{4-5v-v(5+v)}{5+v} \\ \text{or,}~~ x'\frac{dv}{dx'}=\frac{4-5v-5v-v^2}{5+v} \\ \text{or,}~~ x'\frac{dv}{dx'}=\frac{4-10v-v^2}{5+v} \\ \text{or,}~~ \displaystyle\int{\frac{5+v}{v^2+10v-4}~dv}=-\int{\frac{dx'}{x'}} \\ \text{or,}~~ \frac 12\log|v^2+10v-4|=-\log|x'|+\frac 12\log c_1 \\ \text{or,}~~\log\left|\frac{y'^2}{x'^2}+10.\frac{y'}{x'}-4\right|=-2\log|x'|+\log c_1 \\ \text{or,}~~\log\left|\frac{y'^2}{x'^2}+10.\frac{y'}{x'}-4\right|+\log|x'|^{2}=\log c_1 \\ \text{or,}~~ \log\left|\left(\frac{y'^2}{x'^2}+10.\frac{y'}{x'}-4\right) x'^2\right|=\log c_1 \\ \text{or,}~~ \log|y'^2+10x'y'-4x'^2|=\log c_1 \\ \text{or,}~~ y'^2+10x'y'-4x'^2=-c\\~~~[\text{where}~~c_1=-c ;\\~~~ \frac 12 \log c_1 \rightarrow \text{constant of integration.}] \\ \text{or,}~~ (y-23/29)^2+10(x-7/29)(y-23/29)\\~~~-4(x-7/29)^2=-c\\ \therefore y^2-(46/29)y+(23/29)^2+10xy\\~~-(70/29)y-(230/29)x+\frac{1610}{29 \times 29}\\~~-4x^2+\frac{56}{29}x-\frac{4 \times 49}{29^2}=-c \\ \text{or,}~~4x^2-y^2-10xy+6x+4y=c~~\text{(ans.)}$

Alternative way :

$~1.~~\frac{dy}{dx}=\frac{4x-5y+3}{5x+y-2}$

Sol.  $~~~\frac{dy}{dx}=\frac{4x-5y+3}{5x+y-2} \\ \text{or,}~~5x~dy+y~dy-2~dy\\~~~=4x~dx-5y~dx+3~dx \\ \text{or,}~~5(x~dy+y~dx)+y~dy-2~dy\\~~~=4x~dx+3~dx \\~~~~~ \text{Integrating we get,} \\ \text{or,}~~5\displaystyle\int{d(xy)}+\int{y~dy}-2\int{dy}\\~~~=4\int{x~dx}+3\int{dx} \\ \text{or,}~~5xy+\frac{y^2}{2}-2y=4\times \frac{x^2}{2}+3x+\frac{c}{2} \\ \text{or,}~~ 10xy+y^2-4y=4x^2+6x+c \\ \text{or,}~~0=4x^2-10xy-y^2+6x-4y+c \\ \text{or,}~~4x^2-10xy-y^2+6x-4y+c=0~~\text{(ans.)}$

Note : $~~\frac c2 \rightarrow \text{constant of integration.}$

$~2.~~\frac{dy}{dx}=\frac{3x+4y+1}{-4x+2y-3}$

Sol. $~~\frac{dy}{dx}=\frac{3x+4y+1}{-4x+2y-3} \\ \text{or,}~~ -4x~dy+2y~dy-3~dy=3x~dx\\~~~~~~~~~~~~~+4y~dx+dx \\ \text{or,}~~2y~dy-3~dy=3x~dx\\~~~~~~~~~~~~~+4(y~dx+x~dy)+dx \\~~~~~~~ \text{Integrating we get,}~~\\ \\ \text{or,}~~ 2\displaystyle\int y~dy-3\int{dy}=3\displaystyle\int{x~dx}\\~~~~~~~~~~~~~+4\displaystyle\int{d(xy)}+\int{dx} \\ \text{or,}~~ 2\cdot\frac{y^2}{2}-3y=3\cdot \frac{x^2}{2}+4xy+x+\frac c2~~[*] \\ \text{or,}~~ 2(y^2-3y)=3x^2+8xy+2x+c \\~~~[ \text{Multiplying both sides by }~2] \\ \text{or,}~~0=3x^2+8xy-2(y^2-3y)+2x+c \\ \text{or,}~~3x^2+8xy-2y^2+2x+6y+c=0~~\text{(ans.)} $

Note[*] : $~~\frac c2 \rightarrow \text{constant of integration.}$

$~3.~~(6x+5y-2)~dx+(5x-3y+2)~dy=0$

Sol. $~~~(6x+5y-2)~dx+(5x-3y+2)~dy=0 \\ \text{or,}~~ 6x~dx+5y~dx-2~dx+5x~dy\\~~~~~-3y~dy+2~dy=0 \\ \text{or,}~~6x~dx+5(y~dx+x~dy)-2~dx\\~~~~~-3y~dy+2~dy=0 \\ ~~~~~~\text{Integrating we get,}~~ \\ \text{or,}~~6\displaystyle\int{x~dx+5\int{d(xy)}}-2\displaystyle\int{dx}\\~~~~~-3\displaystyle\int{y~dy}+2\int{dy}=0 \\ \text{or,}~~ 6 \times \frac{x^2}{2}+5xy-2x\\~~~~~-3\times \frac{y^2}{2}+2y=\frac{c_1}{2} \\ ~~~~~[\because~ y~dx+x~dy=d(xy)] \\ \text{or,}~~2\left(3x^2+5xy-2x-\frac 32 y^2+2y\right)=2 \times \frac{c_1}{2} \\ \text{or,}~~ 6x^2+10xy-4x-3y^2\\~~~~~+4y=-c~~[\text{taking}~c_1=-c]\\ \text{or,}~~ 6x^2+10xy-3y^2-4x\\~~~~~+4y+c=0~~\text{(ans.)}$

$~4.~~\frac{dy}{dx}=\frac{x+y}{2(x+y)+3}$

Sol. $~~~\frac{dy}{dx}=\frac{x+y}{2(x+y)+3}\rightarrow(1)$

Let $~~~z=x+y \\ \text{or,}~~\frac{dz}{dx}=1+\frac{dy}{dx} \rightarrow(2)$

By $~(1),(2)~$ we get,

$~~~\frac{dz}{dx}-1=\frac{z}{2z+3} \\ \text{or,}~~\frac{dz}{dx}=1+\frac{z}{2z+3} \\ \text{or,}~~\frac{dz}{dx}=\frac{2z+3+z}{2z+3} \\ \text{or,}~~ \frac{dz}{dx}=\frac{3z+3}{2z+3}\\ \text{or,}~~\frac{2z+3}{3z+3}~dz=dx \\ \text{or,}~~ \frac{2(z+1)+1}{3(z+1)}~dz=dx \\ \therefore \frac 23~dz+\frac 13 \times \frac{dz}{z+1}=dx \\ \text{or,}~~\frac 23 \displaystyle\int{dz}+\frac 13\int{\frac{dz}{z+1}}=\displaystyle\int{dx} \\ \text{or,}~~\frac 23z+\frac 13\log|z+1|=x+c_1 \\ \text{or,}~~2z+\log|z+1|-3x=3c_1 \\ \text{or,}~~2(x+y)-3x+\log|x+y+1|=c \\ \text{or,}~~ -x+2y+\log|x+y+1|=c \\ \text{or,}~~ \log|x+y+1|=x-2y+c~~\text{(ans.)}$

Note : $~~c_1 \rightarrow \text{constant of integration},~~3c_1=c.$

$~5.~~\frac{dy}{dx}=\frac{x+y+1}{x+y-1},\\~~~\text{given}~~y=1,~~\text{when}~~x=1.$

Sol. $~~~\frac{dy}{dx}=\frac{x+y+1}{x+y-1}\rightarrow(1)$

Let $~~z=x+y \\ \text{or,}~~\frac{dz}{dx}=1+\frac{dy}{dx} \rightarrow(2)$

Then by $~(1),~(2)~$ we get,

$~~\frac{dz}{dx}-1=\frac{z+1}{z-1} \\ \text{or,}~~ \frac{dz}{dx}=\frac{z+1}{z-1}+1 \\ \text{or,}~~ \frac{dz}{dx}=\frac{z+1+z-1}{z-1} \\ \text{or,}~~\frac{dz}{dx}=\frac{2z}{z-1} \\ \text{or,}~~\frac{z-1}{2z}~dz=dx \\ \text{or,}~~\frac 12\left(1-\frac 1z\right)~dz=dx \\ \text{or,}~~ \frac 12 \displaystyle\int{dz}-\frac 12 \int{\frac{dz}{z}}=\int{dx}+c_1 \\ \text{or,}~~\frac{z}{2}-\frac 12\log|z|=x+c_1 \\ \text{or,}~~ \frac 12(x+y)-\frac 12\log|x+y|=x+c_1 \\ \text{or,}~~(x+y)-\log|x+y|=2x+2c_1 \\ \text{or,}~~ y-\log|x+y|=x+2c_1 \rightarrow(3)$

Since when $~x=1,~y=1,~~$ we get from $~(3),$

$~~1-\log|1+1|=1+2c_1 \\ \text{or,}~~2c_1=-\log 2$

Now, putting the value of $~2c_1~$ in $~(3)~$ we get,

$~~y-\log|x+y|=x-\log 2 \\ \text{or,}~~\log 2=x+\log|x+y|-y \\ \text{or,}~~ x-y+\log|x+y|=\log 2~~\text{(ans.)}$

Note : $~~c_1 \rightarrow \text{constant of integration.}$

$~6.~~(2x-2y+5)~dy=(x-y+3)~dx$

Sol. $~~~(2x-2y+5)~dy=(x-y+3)~dx \\ \text{or,}~~ \frac{dy}{dx}=\frac{x-y+3}{2(x-y)+5}\rightarrow(1)$

Let $~~~~z=x-y \\ \text{or,}~~\frac{dz}{dx}=1-\frac{dy}{dx} \\ \text{or,}~~\frac{dy}{dx}=1-\frac{dz}{dx}\rightarrow(2)$

By $~(1),(2)~$ we get,

$~~~1-\frac{dz}{dx}=\frac{z+3}{2z+5} \\ \text{or,}~~ 1-\frac{z+3}{2z+5}=\frac{dz}{dx} \\ \text{or,}~~ \frac{dz}{dx}=\frac{2z+5-(z+3)}{2z+5} \\ \text{or,}~~\frac{dz}{dx}=\frac{2z+5-z-3}{2z+5} \\ \text{or,}~~ \frac{dz}{dx}=\frac{z+2}{2z+5} \\ \text{or,}~~\frac{2z+5}{z+2}~dz=dx \\ \text{or,}~~\displaystyle\int{\left[\frac{2(z+2)+1}{z+2}\right]}~dz=\displaystyle\int{dx} \\ \text{or,}~~ 2\displaystyle\int{dz}+\int{\frac{dz}{z+2}}=x+c \\ \text{or,}~~ 2z+\log|z+2|=x+c \\ \text{or,}~~2(x-y)-x+\log|x-y+2|=c \\ \text{or,}~~ (x-2y)+\log|x-y+2|=c~~\text{(ans.)}$

Note : $~~c \rightarrow \text{constant of integration.}$