Differential Equations

Given,  $\frac{dy}{dx}=ytanx$

$\Rightarrow \frac{dy}{y}=tanxdx$

Integrating both sides we get,

$\begin{array}{l}{\displaystyle \int \frac{dy}{y}}={\displaystyle \int tanxdx}\\ \Rightarrow logy=log\left|secx\right|+logc\\ \Rightarrow logy=log\left|csecx\right|\\ \Rightarrow y=c_{1}secx\left(where,c_1=\pm c\right)\end{array}$

As,  $y=1,at,x=0$ we have,

$1=c_{1}sec\left(0\right)=c\Rightarrow c=1$

$\therefore$ The required particular solution is $y=secx$ .

Given, D.E. is

$\begin{array}{l}cos\frac{dy}{dx}=a\\ \Rightarrow \frac{dy}{dx}=co{s}^{-1}\left(a\right)\\ \Rightarrow dy=co{s}^{-1}\left(a\right)dx\end{array}$

Integrating both sides,

$\begin{array}{l}{\displaystyle \int dy={\displaystyle \int co{s}^{-1}\left(a\right)dx}}\\ \Rightarrow y=co{s}^{-1}\left(a\right)*x+c\\ \Rightarrow y=xco{s}^{-1}\left(a\right)dx\end{array}$

Given,  $y=1,atx=0$

Then,  $1=0co{s}^{-1}\left(a\right)+c$

$\Rightarrow c=1$

$\therefore$ The required particular solution is

The given D.E. is

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

Integrating both sides,

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

Let, $\frac{1}{x\left(x-1\right)\left(x+1\right)}=\frac{A}{x}+\frac{B}{x-1}+\frac{c}{x+1}$

$\begin{array}{l}1=A\left(x-1\right)\left(x+1\right)+B\left(x\right)\left(x+1\right)+C\left(x\right)\left(x-1\right)\\ =A\left(x^2-1\right)+B{x}^2+Bx+C{x}^2-Cx\\ =A{x}^2-A+B{x}^2+Bx+C{x}^2-Cx\\ =\left(A+B+C\right)x^2+\left(B-C\right)x-A\end{array}$

Comparing the coefficient,

$\begin{array}{l}-A=1\\ \Rightarrow A=-1------------(1)\\ \Rightarrow A+B+C=0---------(2)\\ \Rightarrow B-C=0\\ \Rightarrow B=C-------------(3)\end{array}$

Putting equation (1) & (2) in (1) we get,

$\begin{array}{l}-1+B+B=0\\ \Rightarrow -1+2B=0\\ \Rightarrow B=\frac{1}{2}=C\end{array}$

So, $\frac{1}{x\left(x-1\right)\left(x+1\right)}=-\frac{1}{x}+\frac{\frac{1}{2}}{x-1}+\frac{\frac{1}{2}}{x+1}$

$=-\frac{1}{x}+\frac{1}{2\left(x-1\right)}+\frac{1}{2\left(x+1\right)}$

Integrating becomes,

$\begin{array}{l}y={\displaystyle \int -\frac{1}{x}dx+{\displaystyle \int \frac{1}{2\left(x-1\right)}dx+{\displaystyle \int \frac{1}{2\left(x+1\right)}dx+c}}}\\ =-log\left(x\right)+\frac{1}{2}log\left(x-1\right)+\frac{1}{2}log\left(x+1\right)+c\\ =\frac{1}{2}\left[-2log\left(x\right)+log\left(x-1\right)+log\left(x+1\right)\right]+c\\ =\frac{1}{2}\left[-log{x}^2+log\left(x+1\right)\left(x-1\right)\right]+c\\ =\frac{1}{2}log\frac{x^2-1}{x^2}+c\end{array}$

Given, $y=0whenx=2.$

Then, $0=\frac{1}{2}log\frac{2^2-1}{2^2}+c$

$\begin{array}{l}\Rightarrow 0=\frac{1}{2}log\frac{3}{4}+c\\ \Rightarrow c=-\frac{1}{2}log\frac{3}{4}\end{array}$

$\therefore$ The required particular solution is

$y=\frac{1}{2}log\frac{x^2-1}{x^2}-\frac{1}{2}log\frac{3}{4}$

The given D.E is $\left(x^3+x^2+x+1\right)\frac{dy}{dx}=2x^2+x$

$\begin{array}{l}\Rightarrow dy=\left(\frac{2x^2+x}{x^3+x^2+x+1}\right)dx\\ \Rightarrow dy=\frac{2x^2+x}{x^2\left(x+1\right)+\left(x+1\right)}dx=\frac{2x^2+x}{\left(x+1\right)\left(x^2+1\right)}dx\end{array}$

Integrating both sides we get,

${\displaystyle \int dy={\displaystyle \int \frac{2x^2+x}{\left(x+1\right)\left(x^2+1\right)}dx}}$

Let, $\frac{2x^2+x}{\left(x+1\right)\left(x^2+1\right)}=\frac{A}{x+1}+\frac{Bx+c}{x^2+1}$

$\begin{array}{l}\Rightarrow 2x^2+2=A\left(x^2+1\right)+\left(Bx+c\right)\left(x+1\right)\\ =A{x}^2+A+B{x}^2+Bx+Cx+C\\ =\left(A+B\right)x^2+\left(B+C\right)x^2+\left(A+C\right)\end{array}$

Comparing the co-efficient we get,

$\begin{array}{l}A+B=2------(1)\\ B+C=1------(2)\\ A+C=0------(3)\end{array}$

Subtracting equation (1) – (2), we get

$\begin{array}{l}A+B-\left(B+C\right)=2-1\\ \to A-C=1\end{array}$

But from equation (3) $A=-C$ so, we get,

$\begin{array}{l}A\left(-C\right)-C=1\\ \to -2C=1\\ \to C=-\frac{1}{2}\\ \&A=-\left(-\frac{1}{2}\right)=\frac{1}{2}\end{array}$

And putting value of A in equation (1),

$\begin{array}{l}\frac{1}{2}+B=2\\ \Rightarrow B=2-\frac{1}{2}=\frac{4-1}{2}=\frac{3}{2}\end{array}$

Putting value of A,B and C in

$\begin{array}{l}\frac{2x^2+x}{\left(x+1\right)\left(x^2+1\right)}=\frac{\frac{1}{2}}{x+1}+\frac{\frac{3}{2}x-\frac{1}{2}}{x^2+1}\\ =\frac{1}{2\left(x+1\right)}+\frac{3}{2}\left(\frac{x}{x^2+1}\right)-\frac{1}{2}\left(\frac{1}{x^2+1}\right)\end{array}$

Hence, the integration becomes

$\begin{array}{l}{\displaystyle \int dy={\displaystyle \int \frac{1}{2\left(x+1\right)}dx+{\displaystyle \int \frac{3}{4}\left(\frac{2x}{x^2+1}\right)dx-{\displaystyle \int \frac{1}{2}\left(\frac{1}{x^2+1}\right)dx}}}}\\ \Rightarrow y=\frac{1}{2}log\left(x+1\right)+\frac{3}{4}log\left(x^2+1\right)-\frac{1}{2}ta{n}^{-1}\frac{x}{1}+c\end{array}$

Given, At $x=0,y=1$

Then, $1=\frac{1}{2}log1+\frac{3}{4}log1-\frac{1}{2}ta{n}^{-1}\left(0\right)+C$

$\begin{array}{l}\Rightarrow 1=0+0-0+C\left\{\begin{array}{l}?log1=0\\ ta{n}^{-1}0-0\end{array}\right\}\\ \Rightarrow c=1\end{array}$

$\therefore$ The required particular solution is:

$\begin{array}{l}y=\frac{1}{2}log\left(x+1\right)+\frac{3}{4}log\left(x^2+1\right)-\frac{1}{2}ta{n}^{-1}x+1\\ =\frac{1}{4}\left[2log\left(x+1\right)+3log\left(x^2+1\right)\right]-\frac{1}{2}ta{n}^{-1}x+1\\ =\frac{1}{4}\left[log{\left(x+1\right)}^2+log{\left(x^2+1\right)}^3\right]-\frac{1}{2}ta{n}^{-1}x+1\\ =\frac{1}{4}\left[log{\left(x+1\right)}^2{\left(x^2+1\right)}^3\right]-\frac{1}{2}ta{n}^{-1}x+1\end{array}$

Given, $e^{x}tanydx+\left(1-e^x\right)se{c}^{2}ydy=0$

Dividing throughout by $\left(1-e^x\right)tany$ we get,

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

Integrating both sides

$\begin{array}{l}={\displaystyle \int \frac{-e^x}{1-e^x}dx+{\displaystyle \int \frac{se{c}^{2}y}{tany}dy=clogc}}\\ =-log\left|1-e^x\right|+log\left|tany\right|=clogc\\ =log\frac{tany}{1-e^x}=logc\\ =\frac{tany}{1-e^x}=c\end{array}$

$=tany=\left(1-e^x\right)c$ is the general solution.

Given,  $\frac{dy}{dx}=si{n}^{-1}x$

$\Rightarrow dy=si{n}^{-1}xdx$

Integrating

Given,  $x^5\frac{dy}{dx}=-y^5$

$\Rightarrow \frac{dy}{y^5}=-\frac{dx}{x^5}$

Integrating both sides

$\begin{array}{l}{\displaystyle \int \frac{dy}{y^5}=-{\displaystyle \int \frac{dx}{x^5}}}\\ \Rightarrow {\displaystyle \int y^{-5}dy=-{\displaystyle \int x^{-5}dx}}\end{array}$

$\begin{array}{l}\Rightarrow \frac{y^{-5+1}}{\left(-5+1\right)}=-\frac{x^{-5+1}}{\left(-5+1\right)}+c\\ \Rightarrow \frac{1}{-4y^4}=\frac{1}{4x^4}+c\end{array}$

$\Rightarrow \frac{1}{y^4}=\frac{1}{x^4}+4c$ is the general solution.

Given, $ylogydx-xdy=0$

$\begin{array}{l}\Rightarrow ylogydx=xdy\\ \Rightarrow \frac{dy}{ylogy}=\frac{dx}{x}\end{array}$

Integration both sides,

${\displaystyle \int \frac{dy}{ylogy}={\displaystyle \int \frac{dx}{x}}}$

Put log $y=t\Rightarrow \frac{1}{y}=\frac{dt}{dy}\Rightarrow \frac{dy}{y}=dt$

Hence, ${\displaystyle \int \frac{dt}{t}={\displaystyle \int \frac{dx}{x}}}$

$\begin{array}{l}\Rightarrow log\left|t\right|=log\left|x\right|+log\left|c\right|=log\left|xc\right|\\ \Rightarrow t=\pm xc\end{array}$

$\Rightarrow logy=ax$ where $a=\pm c$

$\Rightarrow y=e^{ax}$ is the general solution.

Given,  $\frac{dy}{dx}=\left(1+x^2\right)\left(1+y^2\right)$

$\Rightarrow \frac{dy}{\left(1+y^2\right)}=\left(1+x^2\right)dx$

Integrating both sides

$\begin{array}{l}{\displaystyle \int \frac{dy}{\left(1+y^2\right)}dy={\displaystyle \int \left(x^2+1\right)dx}}\\ \Rightarrow ta{n}^{-1}\frac{y}{1}=\frac{x^3}{3}+x+c\end{array}$

$\Rightarrow ta{n}^{-1}y=\frac{x^3}{3}+x+c$ is the general solution.

Given,  $\left(e^x+e^{-x}\right)dy-\left(e^x-e^{-x}\right)dx=0$