Differentiation (Part-38) | S N De

 

Solve $~[~1-7~]:$

$~1.~~\frac{dy}{dx}=(y-x)^2$

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

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

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

$~~\frac{dz}{dx}+1=z^2 \\ \Rightarrow \frac{dz}{dx}=z^2-1\\ \Rightarrow \frac{1}{z^2-1}~dz=dx \\ \Rightarrow \int{\frac{dz}{z^2-1}}=\int{dx} \\ \Rightarrow \frac 12\log\left|\frac{z-1}{z+1}\right|=x+c_1\\~~~ \left[c_1 \rightarrow \text{Const. of Integration.}\right] \\ \Rightarrow \frac 12\log \left|\frac{y-x-1}{y-x+1}\right|=x+c_1 \\ \Rightarrow \log\left|\frac{y-x-1}{y-x+1}\right|=2x+2c_1 \\ \Rightarrow \left|\frac{y-x-1}{y-x+1}\right|=e^{2x+2c_1} \\ \Rightarrow \frac{(y-x-1)^2}{(y-x+1)^2}=e^{4x}.e^{4c_1} \\ \Rightarrow (y-x-1)^2=(y-x+1)^2 c^2. e^{4x}~~\text{(ans.)}\\~~~~~\text{where}~~e^{4c_1}=c^2$

$~2.~~(x-y)^2~\frac{dy}{dx}=1$

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

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

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

$~~~~~z^2\left(1-\frac{dz}{dx}\right)=1\\ \Rightarrow 1-\frac{dz}{dx}=\frac{1}{z^2}\\ \Rightarrow 1-\frac{1}{z^2}=\frac{dz}{dx}\\ \Rightarrow \frac{z^2-1}{z^2}=\frac{dz}{dx}\\ \Rightarrow dx=\left(\frac{z^2}{z^2-1}\right)~dz \\ \Rightarrow \int{dx}=\int{\left(1+\frac{1}{z^2-1}\right)~dz}\\ \Rightarrow \int{dx}=\int{dz}+\int{\frac{1}{z^2-1}~dz}\\ \Rightarrow x+c_1=z+\frac 12\log\left|\frac{z-1}{z+1}\right| \\ \Rightarrow x+c_1=(x-y)+\frac 12 \log\left|\frac{x-y-1}{x-y+1}\right| \\ \Rightarrow c_1+y=\frac 12\log\left|\frac{x-y-1}{x-y+1}\right|\\ \Rightarrow 2(c_1+y)=\log\left|\frac{x-y-1}{x-y+1}\right|\\ \Rightarrow e^{2c_1+2y}=\left|\frac{x-y-1}{x-y+1}\right| \\ \Rightarrow  e^{4c_1+4y}= \left(\frac{x-y-1}{x-y+1}\right)^2 \\ \Rightarrow e^{4c_1}\cdot e^{4x}=\left(\frac{x-y-1}{x-y+1}\right)^2 \\ \Rightarrow (x-y-1)^2=(x-y+1)^2.e^{4y}.c^2~~\text{(ans.)}\\~~~~[\text{where}~~c=e^{2c_1},~~c_1\rightarrow \text{const. of integration.}]$

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

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

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

So, from $\,(1)~\text{and}~~(2),~$we get,

$~~~~z^2\left(\frac{dz}{dx}-1\right)=a^2 \\ \Rightarrow \frac{dz}{dx}-1=\frac{a^2}{z^2}\\ \Rightarrow \frac{dz}{dx}=\frac{a^2}{z^2}+1 \\ \Rightarrow \frac{dz}{dx}=\frac{a^2+z^2}{z^2}\\ \Rightarrow \frac{z^2}{z^2+a^2}~dz=dx \\ \Rightarrow \left(1-\frac{a^2}{z^2+a^2}\right)~dz=dx \\ \therefore \int{\left(1-\frac{a^2}{z^2+a^2}\right)~dz}=\int{dx} \\ \Rightarrow \int{dz}-a^2\int{\frac{1}{z^2+a^2}~dz}=x+c \\ \Rightarrow z-a^2 \times \frac{1}{a}\tan^{-1} \frac za=x+c  \\ \Rightarrow  x+y-a \tan^{-1}\left(\frac{x+y}{a}\right)=x+c \\ \Rightarrow y=a \tan^{-1}\left(\frac{x+y}{a}\right)+c~~\text{(ans.)}\\~~~c\rightarrow \text{Constant of Integration.}$

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

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

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

So, from $\,(1)~\text{and}~~(2),~$we get,

$~~\frac{dz}{dx}-1=z \\ \Rightarrow \frac{dz}{dx}=z+1 \\ \Rightarrow \frac{1}{z+1}~dz=dx\\ \therefore \int{\frac{1}{z+1}~dz}=\int{dx}\\ \Rightarrow \log|z+1|=x+c_1 \\ \Rightarrow \log|x+y+1|=x+c_1 \\ \Rightarrow  |x+y+1|=e^{x+c_1} \\ \Rightarrow (x+y+1)^2=e^{2x+2c_1}\\ \Rightarrow  (x+y+1)^2=e^{2x}.e^{2c_1} \\ \Rightarrow  (x+y+1)^2=e^{2x}.c^2~~\text{(ans.)}\\~~~~\text{where}~~e^{c_1}=c,~~c_1\rightarrow \text{const. of integration.} $

$~5.~~(x+y+1)~\frac{dy}{dx}=1$

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

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

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

$~~z\left(\frac{dz}{dx}-1\right)=1 \\ \Rightarrow \frac{dz}{dx}=\frac 1z+1 \\ \Rightarrow \frac{dz}{dx}=\frac{1+z}{z}  \\ \Rightarrow \frac{z}{z+1}~dz=dx  \\ \Rightarrow \int{\left(1-\frac{1}{z+1}\right)~dz}=\int{dx}  \\ \Rightarrow \int{dz}-\int{\frac{dz}{z+1}}=\int{dx}+c_1 \\ \Rightarrow z-\log|z+1|=x+c_1  \\ \Rightarrow (x+y+1)-\log|x+y+2|=x+c_1  \\ \Rightarrow y+1-\log|x+y+2|=c_1  \\ \Rightarrow (y+1)-c_1=\log|x+y+2|  \\ \Rightarrow |x+y+2|=e^{(y+1)-c_1}  \\ \Rightarrow (x+y+2)^2=e^{2(y+1)}.e^{-2c_1} \\ \Rightarrow (x+y+2)^2=e^{2y}.e^2.e^{-2c_1}  \\ \Rightarrow (x+y+2)^2=e^{2y}.c^2~~\text{(ans.)}\\~~~~\text{where}~~e^2.e^{-2c_1}=c^2.$

$~6.~~\frac{dy}{dx}=\sqrt{x+y+1}$

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

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

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

$~~2z\frac{dz}{dx}-1=z \\ \Rightarrow \frac{dz}{dx}=\frac{z+1}{2z} \\ \Rightarrow \frac{2z}{z+1}~dz=dx \\ \Rightarrow 2 \left(1-\frac{1}{z+1}\right)~dz=dx \\ \therefore 2 \int{dz}-2\int{\frac{1}{z+1}~dz}=\int{dx} \\ \Rightarrow 2z-2\log|z+1|=x+c \\ \Rightarrow 2[\sqrt{x+y+1}-\log|\sqrt{x+y+1}+1|]\\~~~~~~~~=x+c~~\text{(ans.)}\\~~~~c\rightarrow \text{const. of integration.}$

$~7.~~\frac{dy}{dx}=\sin(x+y)$

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

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

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

$~~~~\frac{dz}{dx}-1=\sin z \\ \Rightarrow  \frac{dz}{dx}=1+\sin z \\ \Rightarrow \frac{dz}{1+\sin z}=dx \\ \Rightarrow \frac{(1-\sin z)}{(1-\sin z)(1+\sin z)}~dz=dx \\ \Rightarrow  \int{\frac{1-\sin z}{\cos^2z}~dz}=\int{dx}\\ \Rightarrow  \int{\sec^2z~dz}-\int{\sec z\tan z~dz}=\int{dx}+c \\ \Rightarrow \tan z-\sec z=x+c \\ \Rightarrow  \tan(x+y)-\sec(x+y)=x+c~~\text{(ans.)}\\~~~~~c\rightarrow \text{const. of integration.}$