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As per the latest report, the the highest package offered during Sasurie College of Arts and Science Powered by Sunstone placements in 2025 stood at INR 26 LPA. The same is tabulated below:

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

Sol:

$\begin{array}{l}Givendifferentialequationisdy=cosx\left(2-ycosecx\right)dx\\ \Rightarrow \frac{dy}{dx}=cosx\left(2-ycosecx\right)\\ \Rightarrow \frac{dy}{dx}=2cosx-ycosx.cosecx\\ \Rightarrow \frac{dy}{dx}=2cosx-ycotx\\ \Rightarrow \frac{dy}{dx}+ycotx=2cosx\\ \therefore P=cotxandQ=2cosx\\ IntegratingfactorI.F=e^{{\displaystyle \int Pdx}}=e^{{\displaystyle \int cotx.dx}}=e^{logsinx}=sinx\\ \therefore Solutionoftheequationis\\ y\times I.F.={\displaystyle \int Q\times I.F.dx+c}\\ \Rightarrow y.sinx={\displaystyle \int 2cosx.sinxdx+c}\\ \Rightarrow y.sinx={\displaystyle \int sin2xdx+c}\Rightarrow y.sinx=-\frac{1}{2}cos2x+c\\ Putx=\frac{\pi }{2}andy=2,weget\\ 2.sin\frac{\pi }{2}=-\frac{1}{2}cos2.\frac{\pi }{2}+c\\ \Rightarrow 2\left(1\right)=-\frac{1}{2}cos\pi +c\Rightarrow 2=-\frac{1}{2}\left(-1\right)+c\\ \Rightarrow 2=\frac{1}{2}+c\Rightarrow c=2-\frac{1}{2}=\frac{3}{2}\\ Hence,therequiredequationisysinx=-\frac{1}{2}cos2x+\frac{3}{2}.\end{array}$

Yes, Sasurie College of Arts and Science Powered by Sunstone conducts placement drives annually for its BBA course. The key highlights of Sasurie College of Arts and Science Powered by Sunstone placements for the Class of 2025 are tabulated below:

Students seeking admission to the NLU Kolkata BA LLB (H) course are required to fill out the CLAT 2025 application form. Along with the form, students are required to upload the following documents:

• Relevant marksheets
• A front-facing, passport-size, recent photograph with plain background
• Signature of the candidate
• Category certificate (for SC/ST/OBC candidates)
• Relevant certificate issued by competent authority (for PWD candidates)
• Relevant certificate issued by competent authority (for BPL category candidates)

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

Sol:

$\begin{array}{l}Givendifferentialequationis\\ 2\left(y+3\right)-xy.\frac{dy}{dx}=0\\ \Rightarrow xy.\frac{dy}{dx}=2y+6\\ \Rightarrow \left(\frac{y}{2y+6}\right)dy=\frac{dx}{x}\\ \Rightarrow \frac{1}{2}\left(\frac{y}{y+3}\right)dy=\frac{dx}{x}\\ Integratingbothsidesweget\\ \frac{1}{2}{\displaystyle \int \frac{y}{y+3}.dy}={\displaystyle \int \frac{dx}{x}}\\ \Rightarrow \frac{1}{2}{\displaystyle \int \left(1-\frac{3}{y+3}\right)dy}={\displaystyle \int \frac{dx}{x}}\\ \Rightarrow \frac{1}{2}{\displaystyle \int 1.dy-\frac{3}{2}}{\displaystyle \int \frac{1}{y+3}dy}={\displaystyle \int \frac{dx}{x}}\\ \Rightarrow \frac{1}{2}y-\frac{3}{2}log\left|y+3\right|=logx+c\\ Putx=1,y=-2\\ \Rightarrow \frac{1}{2}\left(-2\right)-\frac{3}{2}log\left|-2+3\right|=log\left(1\right)+c\\ \Rightarrow -1-\frac{3}{2}log\left(1\right)=log\left(1\right)+c\\ \Rightarrow -1-0=0+c\left[?log\left(1\right)=0\right]\\ \therefore c=-1\\ \therefore equationis\\ \frac{1}{2}y-\frac{3}{2}log\left|y+3\right|=logx-1\\ \Rightarrow y-3log\left|y+3\right|=2logx-2\\ \Rightarrow y-log\left|{\left(y+3\right)}^3\right|=log{x}^2-2\\ \Rightarrow log\left|{\left(y+3\right)}^3\right|+log{x}^2=y+2\\ \Rightarrow log\left|x^2{\left(y+3\right)}^3\right|=y+2\\ \Rightarrow x^2{\left(y+3\right)}^3=e^{y+2}\\ Hence,therequiredsol\end{array}$

SCMS School of Engineering and Technology admissions are completely based on entrance exam scores. Students should complete Class 12 as the basic eligibility criteria for degree course and graduation for postgraduate courses. The selection of admission is confirmed on entrance exam scores, verification of the original mark sheet, community certificate and other relevant documents. 

Yes! If you plan smartly, BMS graduates can:

• Apply for a Master's in Management (MiM) or an MBA abroad
• Work in international startups or global MNC's
• Do internships with companies that have a global supply chain
• Learn a foreign language and apply for international customer support/BD roles
• Take certifications like Business Analytics, SAP, Digital Marketing or Trade Certification to stand out from the crowd

In short your international career depends more on your skillset, network and post-BMS qualifications.

Pursuing a BBA course from Sasurie College of Arts and Science Powered by Sunstone is a good choice if considering the placement opportunities. The institute offered more than 5,000 dream jobs across 150 job profiles in the recent past. The key highlights of Sasurie College of Arts and Science Powered by Sunstone placements for the Class of 2025 are tabulated below:

Yes, Priyadarshini Institute of Architecture and Design Studies admissions are open. Interested students can visit the official website of the college to apply for the desired programme by filling out the enquiry form and proceeding with the specified steps. Students need to submit valid scores obtained in the relevant entrance test in addition to fulfilling the minimum educational criteria.

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

Sol:

$\begin{array}{l}Givendifferentialequationis\\ \left(x+y\right)\left(dx-dy\right)=dx+dy\\ \Rightarrow \left(x+y\right)dx-\left(x+y\right)dy=dx+dy\\ \Rightarrow -\left(x+y\right)dy-dy=dx-\left(x+y\right)dx\\ \Rightarrow -\left(x+y+1\right)dy=-\left(x+y-1\right)dx\\ \Rightarrow \frac{dy}{dx}=\frac{x+y-1}{x+y+1}\\ Putx+y=z\\ \therefore 1+\frac{dy}{dx}=\frac{dz}{dx}\\ \Rightarrow \frac{dy}{dx}=\frac{dz}{dx}-1\\ So,\frac{dz}{dx}-1=\frac{z-1}{z+1}\\ \Rightarrow \frac{dz}{dx}=\frac{z-1}{z+1}\Rightarrow \frac{dz}{dx}=\frac{z-1+z+1}{z+1}\\ \Rightarrow \frac{dz}{dx}=\frac{2z}{z+1}\Rightarrow \frac{z+1}{z}dz=2.dx\\ Integratingbothsidesweget\\ {\displaystyle \int \frac{z+1}{z}dz}=2{\displaystyle \int dx}\\ \Rightarrow {\displaystyle \int \left(1+\frac{1}{z}\right)dz}=2{\displaystyle \int dx}\\ \Rightarrow z+log\left|z\right|=2x+log\left|c\right|\\ \Rightarrow x+y+log\left|x+y\right|=2x+log\left|c\right|\\ \Rightarrow y+log\left|x+y\right|=x+log\left|c\right|\\ \Rightarrow log\left|x+y\right|=x-y+log\left|c\right|\\ \Rightarrow log\left|x+y\right|-log\left|c\right|=\left(x-y\right)\\ \Rightarrow log\left|\frac{x+y}{c}\right|=\left(x-y\right)\\ \Rightarrow \frac{x+y}{c}=e^{x-y}\\ \therefore x+y=c.e^{x-y}\\ Hence,therequiredsolutionisx+y=c.\end{array}$

Sasurie College of Arts and Science Powered by Sunstone has a dedicated training and placement cell with an excellent placement record. Sunstone has launched a Placement Opportunity Program. The programme offers students access to over 200 assured placement opportunities, or a refund of their first-year tuition fee if placements don't meet the commitment. Check out the key highlights of Sasurie College of Arts and Science Powered by Sunstone placements for the Class of 2025 in the table below:

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

Sol:

$\begin{array}{l}Givenequationis{y}^{2}dx+\left(x^2-xy+y^2\right)dy=0\\ \Rightarrow \frac{dx}{dy}=-\frac{x^2-xy+y^2}{y^2}\\ Putx=vy\left[?Itisa\text{\hspace{0.17em}homogeneous}differentialequation\right]\\ \therefore \frac{dx}{dy}=v+y.\frac{dv}{dy}\\ \therefore v+y.\frac{dv}{dy}=-\left(\frac{y^2-v{y}^2+v^2{y}^2}{y^2}\right)\\ \Rightarrow v+y.\frac{dv}{dy}=-\frac{y^2\left(1-v+v^2\right)}{y^2}\\ \Rightarrow v+y.\frac{dv}{dy}=\left(-1+v-v^2\right)\\ \Rightarrow y.\frac{dv}{dy}=-1+v-v^2-v\\ \Rightarrow y.\frac{dv}{dy}=-1-v^2\Rightarrow \frac{dv}{v^2+1}=-\frac{dy}{y}\\ Integratingbothsidesweget\\ {\displaystyle \int \frac{dv}{v^2+1}}=-{\displaystyle \int \frac{dy}{y}}\\ \Rightarrow ta{n}^{-1}v=-logy+c\Rightarrow ta{n}^{-1}\left(\frac{x}{y}\right)+logy=c\\ Hence,therequiredsolutionista{n}^{-1}\left(\frac{x}{y}\right)+logy=c.\end{array}$