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Yes, listed below are the best affordable Fashion Designing institutes in TamilNadu having tuition fees of less than INR 1 lakh:

Disclaimer: This information is sourced from official website.

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Thegivendifferentialequationis\\ \frac{dy}{dx}=e^{\frac{x^2}{2}}+xy\Rightarrow \frac{dy}{dx}-xy=e^{\frac{x^2}{2}}\\ \text{\hspace{0.17em}Since}itislineardifferentialequationwhereP=-xandQ=e^{\frac{x^2}{2}}\\ \therefore I.F.=e^{{\displaystyle \int P.dx}}=e^{{\displaystyle \int -x.dx}}=e^{\frac{-x^2}{2}}\\ So,thesolutionis\\ y\times I.F.={\displaystyle \int Q\times }I.F.dx+c\Rightarrow y\times e^{\frac{-x^2}{2}}={\displaystyle \int e^{\frac{x^2}{2}}}.e^{\frac{-x^2}{2}}dx+c\\ \Rightarrow y\times e^{\frac{-x^2}{2}}={\displaystyle \int e^0.}dx+c\Rightarrow y.e^{\frac{-x^2}{2}}={\displaystyle \int 1.}dx+c\\ \Rightarrow y.e^{\frac{-x^2}{2}}=x+c\\ \therefore y=\left(x+c\right)e^{\frac{x^2}{2}}\\ Hence,thecorrectoptionis\left(c\right).\end{array}$

Yes, joining the best Fashion Designing colleges in TamilNadu can be worthwhile for you. For that, you need to check its ROI. Listed below are some colleges along with their tuition fees and average placement package:

Disclaimer: This information is sourced from official website.

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Since,theslopeofthetothecurve=x:y\\ \therefore \frac{dy}{dx}=\frac{x}{y}\Rightarrow ydy=xdx\\ Integratingbothsides,wehave\\ {\displaystyle \int ydy={\displaystyle \int xdx}}\\ \Rightarrow \frac{y^2}{2}=\frac{x^2}{2}+c\Rightarrow y^2=x^2+2c\\ \Rightarrow y^2-x^2=2c=kwhichis\text{\hspace{0.17em}rectangular}hyperbola.\\ Hence,thecorrectoptionis\left(d\right).\end{array}$

Yes, there are about 10+ BDes in Fashion Design colleges in Tamil Nadu. Some of them are mentioned below along with their tuition fees:

Disclaimer: This information is sourced from official website.

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Thegivendifferentialequationis\\ \frac{dy}{dx}=2x.e^{x^2-y}\Rightarrow \frac{dy}{dx}=2x.e^{x^2}.e^{-y}\\ \Rightarrow \frac{dy}{e^{-y}}=2x.e^{x^2}dx\\ Integratingbothsides,wehave\\ {\displaystyle \int \frac{dy}{e^{-y}}={\displaystyle \int 2x.e^{x^2}dx}}\\ \Rightarrow {\displaystyle \int e^{y}dy={\displaystyle \int 2x.e^{x^2}dx}}\left[\begin{array}{l}PutinR.H.S.x^2=t\\ \therefore 2xdx=dt\end{array}\right]\\ \Rightarrow {\displaystyle \int e^{y}dy={\displaystyle \int e^{t}dt}}\\ \Rightarrow e^y=e^t+c\Rightarrow e^y=e^{x^2}+c\\ Hence,thecorrectoptionis\left(c\right).\end{array}$

As per popularity basis, listed below are the top Fashion Design colleges in Tamil Nadu along with their tuition fees:

Disclaimer: This information is sourced from official website.

About 140+ top Fashion Design colleges are there in TamilNadu. Students can secure admission after passing Class 12 boards or through entrance exam like CUET, NIFT entrance exam, TANCET, etc. Some of the top Fashion Design colleges in Tamil Nadu include NIFT Chennai, PSGCAS, Sri Krishna Arts and Science College, Hindusthan College of Arts and Science, and many others. Of these, 88 colleges are privately owned, 19 colleges are owned by public/government organisations and public private entities own 10 colleges.

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Thegivenequationisy=Ax+A^3\\ Differentiatingbothsides,weget\frac{dy}{dx}=A\\ Againdifferentiatingbothsides,weget\frac{d^{2}y}{d{x}^2}=0\\ So,theorderofthedifferentialequationis2.\\ Hence,thecorrectoptionis\left(b\right).\end{array}$

CGPA is a broader term used for an overall year's calculation or for all the academic years. SGPA on the other hand is essential for just semester and not for a year. Thus, CGPA is a better metric to assess academic performance year on year and also the overall academic performance.

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Thegivenequationis{x}^2+y^2-2ay=0\dots \left(i\right)\\ Differentiatingw.r.txwehave\\ 2x+2y.\frac{dy}{dx}-2a\frac{dy}{dx}=0\\ \Rightarrow x+y\frac{dy}{dx}-a\frac{dy}{dx}=0\Rightarrow x+\left(y-a\right)\frac{dy}{dx}=0\\ \Rightarrow \left(y-a\right)\frac{dy}{dx}=-x\Rightarrow y-a=\frac{-x}{dy/dx}\\ \Rightarrow a=\frac{-x}{dy/dx}+y\Rightarrow a=\frac{y.\frac{dy}{dx}+x}{\frac{dy}{dx}}\\ Puttingthevalueofaineqn.\left(i\right)weget\\ \Rightarrow x^2+y^2-2y\left[\frac{y.\frac{dy}{dx}+x}{\frac{dy}{dx}}\right]=0\\ \Rightarrow \left(x^2+y^2\right)\frac{dy}{dx}-2y\left(y.\frac{dy}{dx}+x\right)=0\\ \Rightarrow \left(x^2+y^2\right)\frac{dy}{dx}-2y^2\frac{dy}{dx}-2xy=0\\ \Rightarrow \left(x^2+y^2-2y^2\right)\frac{dy}{dx}=2xy\Rightarrow \left(x^2-y^2\right)\frac{dy}{dx}=2xy\\ Hence,thecorrectoptionis\left(a\right).\end{array}$

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Thegivendifferentialequationisx\frac{dy}{dx}+y=e^x\\ \Rightarrow \frac{dy}{dx}+\frac{y}{x}=\frac{e^x}{x}\\ Here,P=\frac{1}{x}andQ=\frac{e^x}{x}\\ \therefore IntegratingfactorI.F.=e^{{\displaystyle \int P.dx}}=e^{{\displaystyle \int \frac{1}{x}.dx}}=e^{log\left|x\right|}=x\\ So,thesolutionis\\ y\times I.F.={\displaystyle \int Q\times }I.F.dx+k\Rightarrow y\times x={\displaystyle \int \frac{e^x}{x}.}xdx+k\\ \Rightarrow y\times x={\displaystyle \int e^x}dx+k\Rightarrow y\times x=e^x+k\\ \therefore y=\frac{e^x}{x}+\frac{k}{x}\\ Hence,thecorrectoptionis\left(a\right).\end{array}$

Providence College of Engineering – PCE selection criteria is primarily entrance-based. Further, appearing for the accepted entrance test, followed by counselling, is also mandatory. Look at the table below to learn about the general selection criteria for Providence College of Engineering courses: