Fill in the blanks for the following (I to xi)

(i) The degree of the differential equation $\frac{d^{2}y}{d{x}^2}+e^{\frac{dy}{d{x}^{}}}+0$ is _________.

(ii) The degree of the differential equation$$ is _________.

(iii) The number of arbitrary constants in the general solution of a differential equation of order three is _________.

(iv) $\frac{dy}{dx}+\frac{y}{x logx}=\frac{1}{x}$ is an equation is of the type _________.

(v) The general solution of a differential equation of the type $\frac{dy}{dx}+P_{1}x=Q_1$ is given by _________.

(vi) The solution of $\frac{xdy}{dx}+2y=x^2$ is _________.

(vii) The solution of $\left(1+x^2\right)\frac{dy}{dx}+2xy-4x^2=0$ is _________.

(viii) The solution of $ydx+\left(x+xy\right)dy=0$ is _________.

(ix) The general solution of $\frac{dy}{dx}+y=sinx$ is _________.

(x) The solution of $cotydx=xdy$ is _________.

(xi) The integrating factor of $\frac{dy}{dx}+y=\frac{1+y}{x}=0$ is _________.

 

This is a Fill in the Blanks Type Questions as classified in NCERT Exemplar

$\begin{array}{l}\left(vii\right)Thegivendifferentialequationis\left(1+x^2\right)\frac{dy}{dx}+2xy-4x^2=0\\ \Rightarrow \frac{dy}{dx}+\frac{2xy}{1+x^2}=\frac{4x^2}{1+x^2}\\ Since,itisalineardifferentialequationwhereP=\frac{2x}{1+x^2}andQ=\frac{4x^2}{1+x^2}\\ I.F.=e^{{\displaystyle \int Pdx}}=e^{{\displaystyle \int \frac{2x}{1+x^2}dx}}=e^{log\left(1+x^2\right)}=1+x^2\\ \therefore Solutionisy*I.F.={\displaystyle \int Q*I.F}.dx+c\\ \Rightarrow y.\left(1+x^2\right)={\displaystyle \int \frac{4x^2}{1+x^2}*\left(1+x^2\right)dx+c}\\ \Rightarrow y.\left(1+x^2\right)={\displaystyle \int 4x^{2}dx+c}\Rightarrow y.\left(1+x^2\right)=\frac{4}{3}{x}^3+c\\ \Rightarrow y=\frac{4}{3}\frac{x^3}{\left(1+x^2\right)}+c{\left(1+x^2\right)}^{-1}\\ Hence,thesolutionisy=\frac{4}{3}\frac{x^3}{\left(1+x^2\right)}+c{\left(1+x^2\right)}^{-1}\\ \left(viii\right)Thegivendifferentialequationisydx+\left(x+xy\right)dy=0\\ \Rightarrow \left(x+xy\right)dy=-ydx\Rightarrow x\left(1+y\right)dy=-ydx\\ \Rightarrow \frac{1+y}{y}dy=-\frac{1}{x}dx\\ Integratingbothsides,weget\\ {\displaystyle \int \frac{1+y}{y}dy=-{\displaystyle \int \frac{1}{x}dx}}\\ \Rightarrow {\displaystyle \int \left(\frac{1}{y}+1\right)dy=-{\displaystyle \int \frac{1}{x}dx}}\\ \Rightarrow logy+y=-logx+logc\\ \Rightarrow logy+logx+log{e}^y=logc\\ \Rightarrow log\left(xy.e^y\right)=logc\\ \Rightarrow xy=c.e^{-y}\\ Hence,thesolutionisxy=c.e^{-y}\end{array}$