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An MBA essay is a short written answer you give during the MBA admission process. It helps the university know you better. You can talk about your career goals, your past experiences, and why you want to do an MBA.

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Giventhat\frac{dy}{dx}=cos\left(x+y\right)+sin\left(x+y\right)\\ Putx+y=v,ondifferentiatingw.r.t.x,weget,\\ 1+\frac{dy}{dx}=\frac{dv}{dx}\\ \therefore \frac{dy}{dx}=\frac{dv}{dx}-1\\ \therefore \frac{dv}{dx}-1=cosv+sinv\\ \Rightarrow \frac{dv}{dx}=cosv+sinv+1\\ \Rightarrow \frac{dv}{cosv+sinv+1}=dx\\ Integratingbothsides,wehave\\ {\displaystyle \int \frac{dv}{cosv+sinv+1}={\displaystyle \int 1.dx}}\\ \Rightarrow {\displaystyle \int \frac{dv}{\left(\frac{1-ta{n}^2\frac{v}{2}}{1+ta{n}^2\frac{v}{2}}+\frac{2tan\frac{v}{2}}{1+ta{n}^2\frac{v}{2}}+1\right)}={\displaystyle \int 1.dx}}\\ \Rightarrow {\displaystyle \int \frac{1+ta{n}^2\frac{v}{2}}{\left(1-ta{n}^2\frac{v}{2}+2tan\frac{v}{2}+1\right)}dv={\displaystyle \int 1.dx}}\\ \Rightarrow {\displaystyle \int \frac{se{c}^2\frac{v}{2}}{2+2tan\frac{v}{2}}dv={\displaystyle \int 1.dx}}\\ Put2+2tan\frac{v}{2}=t\\ 2.\frac{1}{2}se{c}^2\frac{v}{2}dv=dt\Rightarrow se{c}^2\frac{v}{2}dv=dt\\ \Rightarrow {\displaystyle \int \frac{dt}{2}}={\displaystyle \int 1.dx}\\ \Rightarrow log\left|t\right|=x+c\\ \Rightarrow log\left|2+2tan\frac{v}{2}\right|=x+c\\ \Rightarrow log\left|2+2tan\left(\frac{x+y}{2}\right)\right|=x+c\\ \Rightarrow log2\left[1+tan\left(\frac{x+y}{2}\right)\right]=x+c\\ \Rightarrow log2+log\left[1+tan\left(\frac{x+y}{2}\right)\right]=x+c\\ \Rightarrow log\left[1+tan\left(\frac{x+y}{2}\right)\right]=x+c-log2\\ Hence,therequiredsolutionislog\left[1+tan\left(\frac{x+y}{2}\right)\right]=x+K\left[c-log2=K\right]\end{array}$

According to the MBA admissions procedure of the Surana College Post Graduate Departments, admission is offered based on the scores secured in the accepted entrance test. Students who are interested in gaining admission for the MBA can apply through the official website. The applicants have to provide the supporting documents and the applicable admission fee to finalise their seat.

Vidyavardhini College of Engineering and Technology Vasai Road offers admission in UG and PG level courses on the basis of minimum aggregate in Class 12 and graduation. Furthermore, candidates must secure valid scores in accepted entrance exams such as HSC/ CET/ JEE or other equivalent national-level exams. 

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Giventhat\left(1+tany\right)\left(dx-dy\right)+2xdy=0\\ \Rightarrow \left(1+tany\right)dx-\left(1+tany\right)dy+2xdy=0\\ \Rightarrow \left(1+tany\right)dx-\left(1+tany-2x\right)dy=0\\ \Rightarrow \left(1+tany\right)\frac{dx}{dy}=\left(1+tany-2x\right)\Rightarrow \frac{dx}{dy}=\frac{1+tany-2x}{1+tany}\\ \Rightarrow \frac{dx}{dy}=1-\frac{2x}{1+tany}\Rightarrow \frac{dx}{dy}+\frac{2x}{1+tany}=1\\ Here,P=\frac{2}{1+tany}andQ=1\\ IntegratingfactorI.F=e^{{\displaystyle \int Pdy}}=e^{{\displaystyle \int \frac{2}{1+tany}.dy}}=e^{{\displaystyle \int \frac{2cosy}{siny+cosy}.dy}}\\ =e^{{\displaystyle \int \frac{siny+cosy-siny+cosy}{siny+cosy}.dy}}=e^{{\displaystyle \int \left(1+\frac{cosy-siny}{siny+cosy}\right).dy}}\\ =e^{{\displaystyle \int 1.dy}}.e^{{\displaystyle \int \frac{cosy-siny}{siny+cosy}.dy}}=e^y.e^{log\left(siny+cosy\right)}=e^y.\left(siny+cosy\right)\\ \therefore Solutionoftheequationis\\ x\times I.F.={\displaystyle \int Q\times I.F.dy+c}\\ \Rightarrow x.e^y.\left(siny+cosy\right)={\displaystyle \int 1.e^y.\left(siny+cosy\right)dy+c}\\ \Rightarrow x.e^y.\left(siny+cosy\right)=e^y.siny+c\left[?{\displaystyle \int e^x\left[f\left(x\right)+f^{\text{'}}\left(x\right)\right]dx=e^{x}f\left(x\right)+c}\right]\\ \Rightarrow x\left(siny+cosy\right)=siny+c.e^{-y}\\ Hence,therequiredsolutionisx\left(siny+cosy\right)=siny+c.e^{-y}\end{array}$

MGSU Bikaner offers MSc across various specialisations such as, Computer Science, Environmental Science, Microbiology, Cyber Security, Yoga Studies and Therapy Management, Biotechnology, and others. The seat intake varies depending upon the specialisation chosen by the candidate. For some courses it is 20, however, for others its 40. Students must check the university website to know more about the seat intake, admission procedure, etc. for their desired programme.

Hi, the table below shows the top private Engineering colleges accepting Andhra Pradesh EAMCET score in Andhra Pradesh along with their total tuition fees:

Disclaimer: This information is sourced from the official website/ sanctioning body and may vary.

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Thegivendifferentialequationisy+\frac{d}{dx}\left(xy\right)=x\left(sinx+logx\right)\\ \Rightarrow y+x.\frac{dy}{dx}+y=x\left(sinx+logx\right)\\ \Rightarrow x\frac{dy}{dx}=x\left(sinx+logx\right)-2y\\ \Rightarrow \frac{dy}{dx}=\left(sinx+logx\right)-\frac{2y}{x}\\ \Rightarrow \frac{dy}{dx}+\frac{2}{x}y=\left(sinx+logx\right)\\ Here,P=\frac{2}{x}andQ=\left(sinx+logx\right)\\ IntegratingfactorI.F=e^{{\displaystyle \int Pdx}}=e^{{\displaystyle \int \frac{2}{x}.dx}}=e^{2logx}=e^{log{x}^2}=x^2\\ \therefore Solutionoftheequationis\\ y\times I.F.={\displaystyle \int Q\times I.F.dx+c}\\ \Rightarrow y.x^2={\displaystyle \int \left(sinx+logx\right)x^{2}dx+c}\dots \left(1\right)\\ LetI={\displaystyle \int \left(sinx+logx\right)x^{2}dx}\\ ={\displaystyle \int \underset{I}{x^2}}\underset{II}{sinx}dx+{\displaystyle \int \underset{II}{x^2}}\underset{I}{logx}dx\\ =\left[x^2.{\displaystyle \int sinxdx-{\displaystyle \int \left(D\left(x^2\right).{\displaystyle \int sinxdx}\right)dx}}\right]+\left[logx.{\displaystyle \int x^{2}dx-{\displaystyle \int \left(D\left(logx\right).{\displaystyle \int x^{2}dx}\right)dx}}\right]\\ =\left[x^2\left(-cosx\right)-2{\displaystyle \int -xcosxdx}\right]+\left[logx.\frac{x^3}{3}-{\displaystyle \int \frac{1}{x}.\frac{x^3}{3}dx}\right]\\ =\left[-x^{2}cosx+2\left(xsinx-{\displaystyle \int 1.sinxdx}\right)\right]+\left[\frac{x^3}{3}logx-\frac{1}{3}{\displaystyle \int x^{2}dx}\right]\\ =-x^{2}cosx+2xsinx+2cosx+\frac{x^3}{3}logx-\frac{1}{9}{x}^3\\ Nowfromeq\left(1\right)weget,\\ y.x^2=-x^{2}cosx+2xsinx+2cosx+\frac{x^3}{3}logx-\frac{1}{9}{x}^3+c\\ \therefore y=-cosx+\frac{2sinx}{x}+\frac{2cosx}{x^2}+\frac{xlogx}{3}-\frac{1}{9}x+c.x^{-2}\\ Hence,therequiredsolutionis\\ y=-cosx+\frac{2sinx}{x}+\frac{2cosx}{x^2}+\frac{xlogx}{3}-\frac{1}{9}x+c.x^{-2}\end{array}$