Differentiation (Part-38) | S N De

Solve:

 $~7.~~\frac{dy}{dx}=\frac{x-y+1}{2x-2y+3}$

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

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,

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

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

$~8.~~(x+y+1)~dx+(2x+2y-1)~dy=0$

Sol. $~~~(x+y+1)~dx+(2x+2y-1)~dy=0 \\ \text{or,}~~\frac{dy}{dx}=-\frac{x+y+1}{2(x+y)-1}\rightarrow(1)$

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

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

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

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

$~9.~~(2x+4y+3)~\frac{dy}{dx}=2y+x+1$

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

Let $~~~~z=2y+x \\ \text{or,}~~\frac{dz}{dx}=2\frac{dy}{dx}+1 \\  \text{or,}~~\frac{dy}{dx}=\frac 12\left(\frac{dz}{dx}-1\right)\rightarrow(2)$

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

$~~~~\frac 12\left(\frac{dz}{dx}-1\right)=\frac{z+1}{2z+3} \\ \text{or,}~~\frac{dz}{dx}-1=\frac{2z+2}{2z+3}\\ \text{or,}~~\frac{dz}{dx}=\frac{2z+2}{2z+3}+1 \\ \text{or,}~~\frac{dz}{dx}=\frac{2z+2+2z+3}{2z+3} \\ \text{or,}~~ \frac{dz}{dx}=\frac{4z+5}{2z+3} \\ \text{or,}~~ \frac{2z+3}{4z+5}~dz=dx \\ \text{or,}~~\displaystyle\int{\left[\frac 12 \cdot \frac{(4z+5)+1}{4z+5}\right]}~dz=\int{dx} \\ \text{or,}~~ \frac 12 \displaystyle\int{dz}+\frac 12\int{\frac{dz}{4z+5}}=x+c_1 \\ \text{or,}~~ \frac z2+\frac 12 .\frac 14 \displaystyle\int{\frac{4}{4z+5}}~dz x+c_1 \\ \text{or,}~~ \frac 12z+\frac 18\log|4z+5|=x+c_1 \\ \text{or,}~~ \frac 12(2y+x)+\frac 18\log|4(2y+x)+5|=x+c_1 \\ \text{or,}~~ 4(2y+x)+\log|8y+4x+5|=8x+8c_1 \\ \text{or,}~~ \log|4x+8y+5|=4x-8y+c \\ \text{or,}~~ \log|4x+8y+5|=4(x-2y)+c~~\text{(ans.)}$

Note : $~~c_1 \rightarrow \text{constant of integration.}\\~~8c_1=c.$

Find the particular solutions of the following equations: 

$~10.~~\frac{dy}{dx}=\frac{1-3x-3y}{2x+2y};\\~~~\text{given}~~y=1,~\text{when}~~x=-1.$

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

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

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

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

Since for $~x=-1,~~y=1$, we have from $~(3),$

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

Putting the value of $~c_1~$ in $~(3),~$ we get,

$~~-2x-2y-2\log|1-x-y|=x+1 \\ \text{or,}~~2x+2y+2\log|1-x-y|=-x-1 \\ \text{or,}~~3x+2y+1+2\log|1-x-y|=0~~\text{(ans.)}$

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

$~11.~~(6x-4y+3)~dx-(3x-2y+1)~dy=0 ;\\~~~\text{given}~~y=4,~~\text{when}~~x=2$

Sol. $~~(6x-4y+3)~dx-(3x-2y+1)~dy=0 \\ \text{or,}~~(6x-4y+3)~dx=(3x-2y+1)~dy \\ \text{or,}~~ \frac{dy}{dx}=\frac{2(3x-2y)+3}{3x-2y+1}\rightarrow(1)$

Let $~~3x-2y=z \\ \text{or,}~~3-2\frac{dy}{dx}=\frac{dz}{dx} \\ \text{or,}~~\frac{dy}{dx}=\frac 12\left(3-\frac{dz}{dx}\right)\rightarrow(2)$

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

$~~~\frac 12\left(3-\frac{dz}{dx} \right)=\frac{2z+3}{z+1} \\ \text{or,}~~3-\frac{dz}{dx} =\frac{4z+6}{z+1} \\ \text{or,}~~\frac{dz}{dx} =3-\frac{4z+6}{z+1} \\ \text{or,}~~\frac{dz}{dx} =\frac{3(z+1)-(4z+6)}{z+1} \\ \text{or,}~~\frac{dz}{dx} =\frac{-(z+3)}{z+1} \\ \text{or,}~~ \frac{z+1}{z+3}~dz=-dx \\ \text{or,}~~\frac{z+3-2}{z+3}~dz=-dx\\ \text{or,}~~\displaystyle\int{\left[1-\frac{2}{z+3}\right]}~dz=-\int{dx} \\ \text{or,}~~\displaystyle \int{dz}-\int{\frac{2}{z+3}}~dz=-\int{dx} \\ \text{or,}~~ z-2\log|z+3|=-x+2\log c \\ \text{or,}~~ (3x-2y)-2\log|3x-2y+3|\\~~~~~=-x+2\log c \\ \text{or,}~~ 3x-2y+x=2\log|3x-2y+3|\\~~~~~+2\log c \\ \text{or,}~~2(2x-y)=2\log|c(3x-2y+3)|\\ \text{or,}~~ e^{2x-y}=c(3x-2y+3)\rightarrow(3)$

For $~~x=2,~~y=4~~$ and so by $~(3),$ we get,

$~~e^{2\times 2-4}=c(3\times 2-2\times 4+3) \\ \text{or,}~~ e^0=c(6-8+3) \\ \text{or,}~~ c=1.$

So, putting the value of $~c~$ in $~(3),~$ we get,

$~~e^{2x-y}=1\times (3x-2y+3) \\ \text{or,}~~ 2x-y=\log|3x-2y+3|~~\text{(ans.)}$ 

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