Differentiation (Part-38) | S N De

 

$~8.~~\frac{dy}{dx}=\tan^2(x+y)$

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

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

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

$~~~~\frac{dz}{dx}-1=\tan^2z \\ \Rightarrow \frac{dz}{dx}=1+\tan^2z  \\ \Rightarrow \frac{dz}{dx}=\sec^2z  \\ \Rightarrow \frac{1}{\sec^2z}~dz=dx  \\ \Rightarrow \frac 12.\int{2\cos^2z~dz}=\int{dx}  \\ \Rightarrow \frac 12 \int{(\cos2z+1)~dz}=x+c_1  \\ \Rightarrow \frac 12\left[\frac{\sin2z}{2}+z\right]=x+c_1  \\ \Rightarrow \frac{\sin 2z}{2}+z=2x+2c_1  \\ \Rightarrow \sin 2z+2z=4x+4c_1 \\ \Rightarrow \sin2(x+y)+2(x+y)=4x+c  \\ \Rightarrow  \sin2(x+y)=2(x-y)+c~~\text{(ans.)}\\~~~~~~\text{where}~~c_1\rightarrow \text{const. of integration},~ c=4c_1.$

$~9.~~\frac{dy}{dx}=\sin(x+y)+\cos(x+y)$

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

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

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

$~~~\frac{dz}{dx}-1=\sin z+\cos z\\ \Rightarrow \frac{dz}{dx}=1+\sin z+\cos z \\ \Rightarrow \int{\frac{dz}{1+\cos z+\sin z}}=\int{dx}\\ \Rightarrow \int{\frac{dz}{2\cos^2\frac z2+2\sin\frac z2\cos\frac z2}}=x+c_1 \\ \Rightarrow\int{\frac{dz}{2\cos^2\frac z2(1+\tan\frac z2)}}=x+c_1 \\ \Rightarrow \int{\frac{\frac 12 \sec^2\frac z2~dz}{1+\tan\frac z2}}=x+c_1 \\ \Rightarrow \int{\frac{\sec^2p~dp}{1+\tan p}}=x+c_1\\~~~~\left[\text{let}~~\frac z2=p\Rightarrow \frac 12 dz=dp\right] \\ \Rightarrow \int{\frac{d(1+\tan p)}{1+\tan p}}=x+c_1 \\~~~[~\because d(1+\tan p)=\sec^2p~dp] \\ \Rightarrow \log\left|1+\tan p\right|=x+c_1 \\ \Rightarrow \left|1+\tan\frac z2\right|=e^{x+c_1} \\ \Rightarrow \left(1+\tan\frac z2\right)^2=e^{2x}.e^{2c_1}\\ \Rightarrow \left(1+\frac{\sin(z/2)}{\cos(z/2)}\right)^2 =c.e^{2x}\\ \Rightarrow \frac{(\cos(z/2)+\sin(z/2))^2}{\cos^2(z/2)}=c.e^{2x} \\ \Rightarrow \frac{\cos^2(z/2)+\sin^2(z/2)+2\sin(z/2)\cos(z/2)}{\cos^2(z/2)}=c.e^{2x} \\ \Rightarrow \frac{2(1+\sin z)}{2\cos^2(z/2)}=c.e^{2x} \\ \Rightarrow 2[1+\sin(x+y)]=c.e^{2x}[1+\cos (x+y)]~~\text{(ans.)}\\~~~[\because 2\cos^2(z/2)=1+\cos z=1+\cos(x+y)]$

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

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

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

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

$~~2-\frac{dz}{dx}=1+e^z \\ \Rightarrow 2-1-e^z=\frac{dz}{dx} \\ \Rightarrow 1-e^z=\frac{dz}{dx} \\ \Rightarrow \frac{1}{1-e^z}~dz=dx \\ \Rightarrow -\int{\frac{-e^{-z}}{e^{-z}-1}~dz}=\int{dx} \\ \Rightarrow -\log|e^{-z}-1|=x+c_1 \\ \Rightarrow \log (|e^{-z}-1|)^{-1}=x+c_1 \\ \Rightarrow \log\left|\frac{1}{e^{-z}-1}\right|=x+c_1 \\ \Rightarrow \frac{1}{e^{-z}-1}=e^x\cdot e^{c_1} \\ \Rightarrow \frac{e^z}{1-e^z}=e^x\cdot e^{c_1} \\ \Rightarrow \frac{e^{2x-y} \cdot e^{-x}}{1-e^{2x-y}}=e^{c_1}\\ \Rightarrow \frac{e^{x-y}}{1-e^{2x-y}}=e^{c_1} \rightarrow (3) $

Now, $~~~\text{given}~~y=2,~~\text{when}~x=2.$

Then by $~(3)~$ we get,

$~~\frac{e^{2-2}}{1-e^{2 \times 2-2}}=e^{c_1} \\ \Rightarrow \frac{1}{1-e^2}=e^{c_1}\rightarrow(4)$

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

$~~\frac{e^{x-y}}{1-e^{2x-y}}=\frac{1}{1-e^2}\\ \therefore \frac{e^{x-y} \cdot e^y}{(1-e^{2x-y})\cdot e^y}=\frac{1}{1-e^2} \\ \Rightarrow \frac{e^x}{e^y-e^{2x}}=\frac{1}{1-e^2} \\ \Rightarrow e^x(1-e^2)=e^y-e^{2x} \\ \Rightarrow -e^x(e^2-1)=-(e^{2x}-e^y) \\ \Rightarrow e^x(e^2-1)=e^{2x}-e^y~~~~\text{(ans.)}$

$~11.~~\frac{dy}{dx}+1=e^{x-y}$

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

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

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

$~~\left(1-\frac{dz}{dx}\right)+1=e^{z} \\ \Rightarrow 2-\frac{dz}{dx}=e^z \\ \Rightarrow \frac{1}{2-e^z}~dz=dx \\ \Rightarrow -\frac 12\int{\frac{-2e^{-z}}{2e^{-z}-1}}~dz=\int{dx} \\ \Rightarrow -\frac 12\log|2e^{-z}-1|=x+c_1 \\ \Rightarrow \log\left|\frac{1}{\sqrt{2e^{-z}-1}}\right|=x+c_1 \\ \Rightarrow \left|\frac{1}{\sqrt{2e^{-z}-1}}\right|=e^x\cdot e^{c_1} \\ \Rightarrow \frac{1}{2e^{-z}-1}=e^{2x}\cdot e^{2c_1} \\ \Rightarrow \frac{e^z}{2-e^z}=e^{2x}\cdot e^{2c_1} \\ \Rightarrow \frac{e^{x-y}}{2-e^{x-y}}=e^{2x}\cdot e^{2c_1} \\ \Rightarrow \frac{e^{x-y} \cdot e^{-2x}}{2-e^{x-y}}=e^{2c_1} \\ \Rightarrow \frac{e^{-(x+y)}}{2-e^{x-y}}=e^{2c_1} \\ \Rightarrow \frac{e^{-(x+y)}\cdot e^y}{(2-e^{x-y})\cdot e^y}=e^{2c_1} \\ \Rightarrow \frac{e^{-x}}{2e^y-e^x}=e^{2c_1} \\ \Rightarrow \frac{1}{e^x(2e^y-e^x)}=e^{2c_1} \\ \Rightarrow e^x(2e^y-e^x)=e^{-2c_1}=c_2,~~\text{(say)} \\ \Rightarrow 2e^y-e^x=c_2e^{-x} \\ \Rightarrow 2e^y=e^x+c_2e^{-x} \\ \therefore e^y=\frac 12 e^x+ce^{-x}~~\text{(ans.)}\\~~~~~\text{where}~~c=\frac 12 c_2,\\~~~~~~c_1 \rightarrow \text{constant of integration.}$

$~12.~~\cos(x+y)~dy=dx$

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

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

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

$~~\cos z\left(\frac{dz}{dx}-1\right)=1 \\ \Rightarrow  \frac{dz}{dx}-1=\frac{1}{\cos z} \\ \Rightarrow  \frac{dz}{dx}=1+\frac{1}{\cos z} \\ \Rightarrow  \frac{dz}{dx}=\frac{\cos z+1}{\cos z} \\ \Rightarrow  \frac{\cos z}{\cos z+1}~dz=dx \\ \Rightarrow  \frac{(1+\cos z)-1}{1+\cos z}~dz=dx \\ \therefore  \int{\left(1-\frac{1}{1+\cos z}\right)~dz}=\int{dx} \\ \Rightarrow  \int{dz}-\int{\frac{1}{2\cos^2(z/2)}}=\int{dx} \\ \Rightarrow  \int{dz}-\frac 12 \int{\sec^2(z/2)~dz}=\int{dx} \\ \Rightarrow  z-\tan(z/2)=x+c\\~~~~~c\rightarrow \text{constant of integration} \\ \Rightarrow  (x+y)-\tan\frac{x+y}{2}=x+c \\ \Rightarrow  y=\tan\frac{x+y}{2}+c~~~\text{(ans.)}$

$~13.~~\tan^{-1}\left(\frac{dy}{dx}\right)=x+y$

Sol. $~~~\tan^{-1}\left(\frac{dy}{dx}\right)=x+y \\ \Rightarrow\frac{dy}{dx}=\tan(x+y)\rightarrow(1)$

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

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

$~~~~\frac{dz}{dx}-1=\tan z \\ \Rightarrow \frac{dz}{dx}=1+\tan z \\ \Rightarrow  \frac{1}{1+\tan z}~dz=dx \\ \therefore  \int{\frac{1}{1+\frac{\sin z}{\cos z}}~dz}=\int{dx} \\ \Rightarrow \int{\frac{\cos z}{\cos z+\sin z}~dz}=\int{dx} \\ \Rightarrow \frac 12 \int{\frac{(\cos z+\sin z)+(\cos z-\sin z)}{\cos z+\sin z}~dz}=\int{dx}\\ \Rightarrow  \frac 12\left[\int{dz}+\int{\frac{d(\cos z+\sin z)}{\cos z+\sin z}}\right]=\int{dx}\\~~~~[\because~d(\cos z+\sin z)=d(\cos z)+d(\sin z)\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=(-\sin z+\cos z)~dz] \\ \Rightarrow  \frac 12 [z+\log|\cos z+\sin z|]=x+\frac{c_1}{2} \\ \Rightarrow z+ \log|\cos z+\sin z|=2x+c_1 \\ \Rightarrow  x+y+\log|\cos(x+y)+\sin(x+y)|\\~~~~~~~~=2x+c_1\\ \Rightarrow  \log|\cos(x+y)+\sin(x+y)|=x-y+c_1 \\ \therefore \cos(x+y)+\sin(x+y)=e^{x-y}\cdot e^{c_1} \\ \Rightarrow  \cos(x+y)+\sin(x+y)=c \cdot e^{x-y}~~\text{(ans.)}\\~~~~\text{where}~~c=e^{c_1};\\~~\frac{c_1}{2}\rightarrow \text{constant of integration.}$

$~14.~~$ Prove that the variables in the equation $~\frac{dy}{dx}=f(ax+by+c)~~$ can be separated by the substitution $~~ax+by+c=z.$

Sol. $~~ax+by+c=z~~\text{(given)}\\ \Rightarrow a+b~\frac{dy}{dx}=\frac{dz}{dx}\\ \Rightarrow b\frac{dy}{dx}=\frac{dz}{dx}-a \\ \Rightarrow \frac{dy}{dx}=\frac 1b\left(\frac{dz}{dx}-a\right)\rightarrow(1)$

Now, $~~\frac{dy}{dx}=f(ax+by+c) \\ \Rightarrow \frac 1b\left(\frac{dz}{dx}-a\right)=f(z)~~[\text{By (1)}] \\ \Rightarrow \frac{dz}{dx}=bf(z)+a \\ \Rightarrow \frac{dz}{bf(z)+a}=dx\rightarrow(2)$

Hence, from $\,(2)\,$ we can conclude that the variables in the equation $~\frac{dy}{dx}=f(ax+by+c)~~$ can be separated by the substitution $~~ax+by+c=z.$