Class 12th

$\begin{array}{l}\left(1+e^{2x}\right)dy+\left(1+y^2\right)e^{x}dx=0\\ \Rightarrow \frac{dy}{1+y^2}+\frac{e^{x}dx}{1+e^{2x}}=0\end{array}$

Integrating both sides, we get:

$\begin{array}{l}ta{n}^{-1}y+{\displaystyle \int \frac{e^{x}dx}{1+e^{2x}}=C..........(1)}\\ Let,e^x=t\Rightarrow e^{2x}=t^2\\ \Rightarrow \frac{d}{dx}\left(e^x\right)=\frac{dt}{dx}\\ \Rightarrow e^x=\frac{dt}{dx}\\ \Rightarrow e^{x}dx=dt\end{array}$

Substituting these values in equation (1), we get:

$\begin{array}{l}ta{n}^{-1}y+{\displaystyle \int \frac{dt}{1+t^2}}=C\\ \Rightarrow ta{n}^{-1}y+ta{n}^{-1}t=C\\ \Rightarrow ta{n}^{-1}y+ta{n}^{-1}(e^x)=C..........(2)\\ Now,y=1,at,x=0\end{array}$

Therefore, equation (2) becomes:

$\begin{array}{l}ta{n}^{-1}1+ta{n}^{-1}1=C\\ \Rightarrow \frac{\pi }{4}+\frac{\pi }{4}=C\\ \Rightarrow C=\frac{\pi }{2}\end{array}$

Substituting $C=\frac{\pi }{2}$ in equation (2), we get:

$ta{n}^{-1}y+ta{n}^{-1}(e^x)=\frac{\pi }{2}$

This is the required solution of the given differential equation.

The differential equation of the given curve is:

$\begin{array}{l}sinxcosydx+cosxsinydy=0\\ \Rightarrow \frac{sinxcosydx+cosxsinydy}{cosxcosy}=0\\ \Rightarrow tanxdx+tanydy=0\end{array}$

Integrating both sides, we get:

$\begin{array}{l}log(secx)+log(secy)=logC\\ log(secx.secy)=logC\\ \Rightarrow secx.secy=C..........(1)\end{array}$

The curve passes through point $\left(0,\frac{\pi }{4}\right)$

$\begin{array}{l}\therefore 1*\surd 2=C\\ \Rightarrow C=\surd 2\end{array}$

On subtracting $C=\surd 2$ in equation (10, we get:

$\begin{array}{l}secx.secy=\surd 2\\ \Rightarrow secx.\frac{1}{cosy}=\surd 2\\ \Rightarrow cosy=\frac{sec\mathrm{x/\surd 2}}{}\end{array}$

Given: Differential equation $\frac{dy}{dx}+\frac{y^2+y+1}{x^2+x+1}=0$

$\begin{array}{l}\frac{dy}{dx}+\frac{y^2+y+1}{x^2+x+1}=0\\ \Rightarrow \frac{dy}{dx}=-\frac{(y^2+y+1)}{x^2+x+1}\\ \Rightarrow \frac{dy}{y^2+y+1}=\frac{-dx}{x^2+x+1}\\ \Rightarrow \frac{dy}{y^2+y+1}+\frac{dx}{x^2+x+1}=0\end{array}$

Integrating both sides,

23. Option (iii) Diamond  is correct since in diamond, the carbon atoms are held together by strong covalent bonds. It is a giant molecule. Thus, it is a solid network.

Kindly go through the solution

22. Option (i) London forces is correct since iodine molecules are nonpolar  and covalent in nature. These molecules are found to be electrically symmetrical and have no dipole moment. The molecules in a crystal lattice of iodine are thus attracted together by weak London forces.

21. Option (ii) a regular arrangement of constituent particles observed over a long distance in the crystal lattice is correct since the regularity of the crystalline lattice creates local environments that are the same and hence crystals exhibit sharp melting point.

20.  (iv) They are anisotropic in nature since amorphous solid shows isotropic properties as they exhibit same values of the properties like refractive index, electrical resistance when measured along different directions.

The equation of a circle in the first quadrant with centre (a, a) and radius (a) which touches the coordinate axes is:

${(x-a)}^2+{(y-a)}^2=a^2..........(1)$

Differentiating equation (1) with respect to x, we get:

$\begin{array}{l}2(x-a)+2(y-a)\frac{dy}{dx}=0\\ \Rightarrow (x-a)+(y-a)y^{\text{'}}=0\\ \Rightarrow x-a+y{y}^{\text{'}}-a{y}^{\text{'}}=0\\ \Rightarrow x+y{y}^{\text{'}}-a(1+y^{\text{'}})=0\\ \Rightarrow a=\frac{x+y{y}^{\text{'}}}{1+y^{\text{'}}}\end{array}$

Substituting the value of a in equation (1), we get:

$\begin{array}{l}{\left[x-\left(\frac{x+y{y}^{\text{'}}}{1+y^{\text{'}}}\right)\right]}^2+{\left[y-\left(\frac{x+y{y}^{\text{'}}}{1+y^{\text{'}}}\right)\right]}^2={\left(\frac{x+y{y}^{\text{'}}}{1+y^{\text{'}}}\right)}^2\\ \Rightarrow {\left[\frac{(x-a)y^{\text{'}}}{(1+y^{\text{'}})}\right]}^2+{\left[\frac{y-x}{1+y^{\text{'}}}\right]}^2={\left[\frac{x+y{y}^{\text{'}}}{1+y^{\text{'}}}\right]}^2\\ \Rightarrow {(x-y)}^2.y^{\text{'}2}+{(x-y)}^2=(x+y{y}^{\text{'}}){}^2\\ \Rightarrow {(x-y)}^2\left[1+(y^{\text{'}}){}^2\right]=(x+y{y}^{\text{'}}){}^2\end{array}$

Hence, the required differential equation of the family of circles is ${(x-y)}^2\left[1+(y^{\text{'}}){}^2\right]=(x+y{y}^{\text{'}}){}^2$

$\Rightarrow \frac{dy}{dx}=\frac{x^3-3x{y}^2}{y^3-3x^{2}y}..........(1)$

This is a homogenous equation. To simplify it, we need to make the substitution as:

$\begin{array}{l}y=vx\\ \Rightarrow \frac{d}{dx}(y)=\frac{d}{dx}(vx)\\ \Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}\end{array}$

Substituting the values of y and $\frac{dv}{dx}$ in equation (1), we get:

$\begin{array}{l}v+x\frac{dv}{dx}=\frac{x^3-3x{(vx)}^2}{{(vx)}^3-3x^2(vx)}\\ \Rightarrow v+x\frac{dv}{dx}=\frac{1-3v^2}{v^3-3v}\\ \Rightarrow x\frac{dv}{dx}=\frac{1-3v^2}{v^3-3v}-v\\ \Rightarrow x\frac{dv}{dx}=\frac{1-3v^2-v(v^3-3v)}{v^3-3v}\\ \Rightarrow x\frac{dv}{dx}=\frac{1-v^4}{v^3-3v}\\ \Rightarrow \left(\frac{v^3-3v}{1-v^4}\right)dv=\frac{dx}{x}\end{array}$

Integrating both sides, we get:

$\begin{array}{l}{\displaystyle \int \left(\frac{v^3-3v}{1-v^4}\right)dv=logx+log{C}^{\text{'}}}.........(2)\\ Now,{\displaystyle \int \left(\frac{v^3-3v}{1-v^4}\right)dv={\displaystyle \int \frac{v^{3}dv}{1-v^4}}}-3{\displaystyle \int \frac{vdv}{1-v^4}}\\ \Rightarrow {\displaystyle \int \left(\frac{v^3-3v}{1-v^4}\right)}dv=I_1-3I_2,Where,I_1={\displaystyle \int \frac{v^{3}dv}{1-v^4}}and{I}_2={\displaystyle \int \frac{vdv}{1-v^4}...........(3)}\end{array}$

$\begin{array}{l}Let,1-v^4=t.\\ \therefore \frac{d}{dv}(1-v^4)=\frac{dt}{dv}\\ \Rightarrow -4v^3=\frac{dt}{dv}\\ \Rightarrow v^{3}dv=-\frac{dt}{4}\\ Now,I_1={\displaystyle \int -\frac{dt}{4}=-logt=-\frac{1}{4}}log(1-v^4)\end{array}$

$\begin{array}{l}And,I_2={\displaystyle \int \frac{vdv}{1-v^4}}={\displaystyle \int \frac{vdv}{1-(v^2){}^2}}\\ Let,v^2=p.\\ \therefore \frac{d}{dv}\left(v^2\right)=\frac{dp}{dv}\\ \Rightarrow 2v=\frac{dp}{dv}\\ \Rightarrow vdv=\frac{p}{2}\\ \Rightarrow I_2=\frac{1}{2}{\displaystyle \int \frac{dp}{1-p^2}=\frac{1}{2*2}log\left|\frac{1+p}{1-p}\right|=\frac{1}{4}log}\left|\frac{1+v^2}{1-v^2}\right|\end{array}$

Substituting the values of $I_1$ and $I_2$ in equation (3), we get:

${\displaystyle \int \left(\frac{v^3-3v}{1-v^4}\right)}dv=-\frac{1}{4}log(1-v^4)-\frac{3}{4}log\left|\frac{1+v^2}{1-v^2}\right|$

Therefore, equation (2) becomes:

$\begin{array}{l}\frac{1}{4}log(1-v^4)-\frac{3}{4}log\left|\frac{1+v^2}{1-v^2}\right|=logx+log{C}^{\text{'}}\\ \Rightarrow -\frac{1}{4}log\left[(1-v^4)\left(\frac{1+v^2}{1-v^2}\right)\right]=log{C}^{\text{'}}x\\ \Rightarrow \frac{{\left(1+v^2\right)}^4}{{\left(1-v^2\right)}^2}={\left(C^{\text{'}}x\right)}^{-4}\\ \Rightarrow \frac{{\left(1+\frac{y^2}{x^2}\right)}^4}{{\left(1-\frac{y^2}{x^2}\right)}^2}=\frac{1}{C^{\text{'}4}{x}^4}\\ \Rightarrow \frac{{\left(x^2+y^2\right)}^4}{x^4{\left(x^2-y^2\right)}^2}=\frac{1}{C^{\text{'}4}{x}^4}\\ \Rightarrow {\left(x^2-y^2\right)}^2=C^{\text{'}4}{\left(x^2+y^2\right)}^4\\ \Rightarrow \left(x^2-y^2\right)=C^{\text{'}2}{\left(x^2+y^2\right)}^2\\ \Rightarrow x^2-y^2=C{\left(x^2+y^2\right)}^2,whereC=C^{\text{'}2}\end{array}$

Hence, the given result is proved.