Differential Equations

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

$Let,e^{\frac{x}{y}}=z$

Differentiating it with respect to y, we get:

$\begin{array}{l}\left(e^{\frac{x}{y}}\right)=\frac{dz}{dy}\\ \Rightarrow e^{\frac{x}{y}}.\frac{d}{dy}\left(\frac{x}{y}\right)=\frac{dz}{dy}\\ \Rightarrow e^{\frac{x}{y}}.\left[\frac{y.\frac{dx}{dy}-x}{y^2}\right]=\frac{dz}{dy}..........(2)\end{array}$

From equation (1) and equation (2), we get:

$\begin{array}{l}\frac{dz}{dy}=1\\ \Rightarrow dz=dy\end{array}$

Integration both sides, we get:

$\begin{array}{l}z=y+C\\ \Rightarrow e^{\frac{x}{y}}y+C\end{array}$

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

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,

Kindly go through the solution

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.

Equation of the given family of curves is  ${(x-a)}^2+2y^2=a^2$

$\begin{array}{l}{(x-a)}^2+2y^2=a^2\\ \Rightarrow x^2+a^2-2ax+2y^2=a^2\\ \Rightarrow 2y^2=2ax-x^2..........(1)\end{array}$

Differentiating with respect to x, we get:

$\begin{array}{l}2y\frac{dy}{dx}=\frac{2a-2x}{2}\\ \Rightarrow \frac{dy}{dx}=\frac{a-x}{2y}\\ \Rightarrow \frac{dy}{dx}=\frac{2a-2x^2}{4xy}..........(2)\end{array}$

From equation (*1), we get:

$2ax=2y^2+x^2$

On substituting this value in equation (3), we get:

$\begin{array}{l}\frac{dy}{dx}=\frac{2y^2+x^2-2x^2}{4xy}\\ \Rightarrow \frac{dy}{dx}=\frac{2y^2-x^2}{4xy}\end{array}$

Hence, the differential equation of the family of curves is given as $\frac{dy}{dx}=\frac{2y^2-x^2}{4xy}$