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The given D.E. is

$\left(1-y^2\right)\frac{dx}{dy}+yx=\propto y\left(-1<y<1\right)$

$\frac{dx}{dy}+\frac{y}{1-y^2}\times x=\frac{\propto y}{1-y^2}$ which is of form

$\begin{array}{l}\frac{dx}{dy}+Px=Q\&P=\frac{y}{1-y^2}\&Q=\frac{\propto y}{1-y^2}\\ {\displaystyle \int pdx={\displaystyle \int \frac{y}{1-y^2}dx=-\frac{1}{2}{\displaystyle \int \frac{-2y}{1-y^{2}dx}}}}\end{array}$

$\begin{array}{l}=-\frac{1}{2}log\left|1-y^2\right|\\ =log{\left[1-y^2\right]}^{-\frac{1}{2}}\end{array}$

$\therefore I.F=e^{{\displaystyle \int Pdx}}=e^{log{\left[1-y^2\right]}^{-\frac{1}{2}}}\\ ={\left[1-y^2\right]}^{-\frac{1}{2}}$

$\therefore$ option (D ) is correct.

15. Option (ii) Low temperature is correct since at sufficiently low temperature, the thermal energy is low, so the intermolecular forces bring the molecules of a substance closer so that they cling to one another and occupy fixed positions. They keep on vibrating about their fixed positions. Such conditions favours the existence of the substance in solid state. 

Yes, after completing an MBA course offered by SRM University AP, students become eligible to work in the public sector of India. This happens because the government sector has many such jobs that require the knowledge and skills gained during an MBA programme. Listed below are some of the prospective government jobs along with their average salaries:

• Bank Probationary Officer (PO): INR 5 LPA to INR 7 LPA
• IBPS PO: INR 3 LPA to INR 4 LPA
• IBPS Clerk: INR 2 LPA to INR 3 LPA
• RBI Assistant/Grade B Officer: INR 6 LPA to INR 8 LPA

Note: The above-mentioned are industry-based average payouts. The actual salary may differ.

The given D.E is

$\begin{array}{l}x\frac{dy}{dx}-y=2x^2\\ \frac{dy}{dx}-\frac{1}{x}y=2x\end{array}$

Which is of form $\frac{dy}{dx}+Py=Q$

So,  $P=-\frac{1}{x}$

$\therefore I.E=e^{{\displaystyle \int Pdx}}=e^{{\displaystyle \int -\frac{1}{x}dx}}=e^{log-x}=e^{log-x^{-1}}=x^{-1}=\frac{1}{x}$

$\therefore$ Option (c) is correct

We know that slope of tangent to the curve is $\frac{dy}{dx}$

$\therefore x+y=\frac{dy}{dx}+5$

$\frac{dy}{dx}-y=x-5$ Which has form $\frac{dy}{dx}+Px=Q$

$where,P=1\&Q=x-5$

$\therefore I.F=e^{{\displaystyle \int Pdx}}=e^{{\displaystyle \int -1dx}}=e^{-x}$

Thus the solution has the form

$\begin{array}{l}y{e}^{-x}={\displaystyle \int \left(x-5\right)e^{-x}dx+c}\\ ={\displaystyle \int x{e}^{-x}dx-{\displaystyle \int 5e^{-x}dx+c}}\\ =y{e}^{-x}=I+5e^{-x}+c\\ where,I={\displaystyle \int x{e}^{-x}dx}\\ =x{\displaystyle \int e^{-x}dx-{\displaystyle \int \frac{d}{dx}x{\displaystyle \int e^{-x}dxdx}}}\\ =-x{e}^{-x}+{\displaystyle \int e^{-x}dx}\\ =-x{e}^{-x}-e^{-x}\end{array}$

$\begin{array}{l}\therefore y{e}^{-x}=-x{e}^{-x}-e^{-x}+5e^{-x}+c\\ =y{e}^{-x}=-x{e}^{-x}+4e^{-x}+c\\ =y=-x+4+\frac{c}{e^{-x}}\\ =y+x=4+c{e}^x\end{array}$

Given, the curve passes through (0,2) so y=2 when x=0

$\begin{array}{l}2+0=4c{e}^0\\ 2-4=c\\ c=-2\end{array}$

$\therefore$ The particular solution is

$y+x=4-2e^x$

SRM University Andhra Pradesh offers an MBA course that runs for a duration of two years. During this period, students are required to pay various fee elements. Some of the prominent fee components included in the fee structure are tuition fees and hostel fees.

Combining all, the total cost to pursue this programme amounts to INR 3.59 lakh. Interested students can refer to the below table to know the breakdown of the total fees:

Note: The above-mentioned fee is as per the official sources. However, it is indicative and subject to change.

We know the slope of tangent to curve is $\frac{dy}{dx}$ .

$\therefore$ $\frac{dy}{dx}=x+y$

$=\frac{dy}{dx}-y=x$ which has form $\frac{dy}{dx}+Py=Q$

So, $P=-1\&Q=x$

$\therefore I.F=e^{{\displaystyle \int Pdx}}=e^{{\displaystyle \int -dx}}=e^{-x}$

Thus the solution is of the form .

$\begin{array}{l}y\times e^{-x}={\displaystyle \int x.e^{-x}dx+c}\\ =x{\displaystyle \int e^{-x}dx-{\displaystyle \int \frac{dx}{dx}{\displaystyle \int e^{-x}dxdx+c}}}\\ =-x{e}^{-x}+{\displaystyle \int e^{-x}dx+c}\\ =y{e}^{-x}=-x{e}^{-x}-e^{-x}+c\\ =y=-x-1+\frac{c}{e^{-x}}\\ =y+x+1=c{e}^x\end{array}$

Given, the curve passes through origin (0,0) i.e, $y=0,when,x=0$

$0+0+1=c{e}^0=c=1$

$\therefore$ Thus, equation of the curve is

$y+x+1=e^x$

Candidates wanting to get admission in BPharm (lateral) course of Motherhood University must meet its eligibility criteria. The aspirants who are DPharm pass from an institution/ university approved by Pharmacy Council of India under section 12 of the Pharmacy Act can apply. Additionally, the minimum aggregate requirement is as per the PCI regulations. 

The given D.E. is

$\frac{dy}{dx}-3ycotx=sin2x$

$\frac{dy}{dx}-3cotx.y=sin2x$ Which is of form $\frac{dy}{dx}+Py=Q$

So, $P=-3cotx\&Q=sin2x$

$\therefore I.F=e^{-{\displaystyle \int Pdx}}=e^{{\displaystyle \int -3cotxdx}}=e^{-3{\displaystyle \int cotxdx}}=e^{-3log\left|cotxdx\right|}=e^{log{\left(sin\right)}^{-3}}=\frac{1}{si{n}^{3}x}$

Thus the solution is of the form.

$\begin{array}{l}y\times \frac{1}{si{n}^{3}x}={\displaystyle \int sin2x.\frac{1}{si{n}^{3}x}dx+c}\\ ={\displaystyle \int \frac{2sinxcosx}{si{n}^{3}x}}dx+c\left\{?sin2x=2sinxcosx\right\}\\ ={\displaystyle \int 2cosecxcotxdx+c}\\ =-2cosecx+c\\ =\frac{y}{si{n}^{3}x}=\frac{-2}{sinx}+c\\ =-2y=-2si{n}^{2}x+csi{n}^{3}x\end{array}$

Given, $y=2,when,x=\frac{\pi }{2}$

$\begin{array}{l}2=-2si{n}^2\frac{1}{2}+csi{n}^3\frac{\pi }{2}\\ =2=-2+c\\ =e=2+2=4\end{array}$

$\therefore$ The particular solution is, $y=-2si{n}^{2}x+4si{n}^{3}x$

The given D.E. is

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

Which is of form 

$\frac{dy}{dx}+Px=Q$

So, $P=\frac{2x}{1+x^2}\&Q=\frac{1}{{\left(1+x^2\right)}^2}$

${\displaystyle \int Pdx={\displaystyle \int \frac{2x}{1+x^2}dx=log\left|1+x^2\right|}}$

$\therefore I.F=e^{{\displaystyle \int Pdx}}=e^{log\left|1+x^2\right|}=1+x^2$

Thus the solution is if form,

$y\times \left(I.F\right)={\displaystyle \int Q}.\left(I.F\right)dx+c$

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

Given, $y=0,when,x=1$

$\begin{array}{l}0=ta{n}^{-1}1+c\\ c=ta{n}^{-1}1=-\frac{\pi }{4}\end{array}$

$\therefore$ The particular solution is

$y\left(1+x^2\right)=ta{n}^{-1}x-\frac{\pi }{4}$

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• Passport Size Photo
• Adhaar card
• Pan card
• Class 10 and 12 Marksheets
• Entrance exam scorecard (if applicable)
• Migration certificate (optional)

The given D.E. is

$\frac{dy}{dx}+2ytanx=sinx$

$\Rightarrow \frac{dy}{dx}+\left(2tanx\right)y=sinx$ Which is of form $\frac{dy}{dx}+Px=Q$

So, $P=2tanx\&Q=sinx$

$\therefore I.F=e^{{\displaystyle \int Pdy}}=e^{{\displaystyle \int 2tanxdx}}=e^{2log\left|secx\right|}=e^{logse{c}^2}=se{c}^2$

Thus the solution is of the form $y\times \left(I.F\right)={\displaystyle \int Q}.\left(I.F\right)dx+c$

$\begin{array}{l}\Rightarrow y.se{c}^{2}x={\displaystyle \int sinx.se{c}^{2}xdx+c}\\ =\frac{sinx}{co{s}^2}dx+c\\ ={\displaystyle \int tanx.secxdx+c}\\ =yse{c}^2=secx+c\\ =y=\frac{1}{secx}+\frac{c}{se{c}^{2}x}=cosx+cco{s}^2\\ =y=cosx+cco{s}^{2}x\end{array}$

Given, $y=0,Whenx=\frac{\pi }{3}$

$\begin{array}{l}=0=cos\frac{\pi }{3}+cco{s}^2\frac{\pi }{3}\left\{\frac{c}{4}=-\frac{1}{2},c=-\frac{4}{2},c=-2\right\}\\ =0=\frac{1}{2}+c{\left(\frac{1}{2}\right)}^2\\ =0=\frac{1}{2}+\frac{c}{4}\end{array}$

C = -2

$\therefore$ The particular solution is

$y=cosx-2co{s}^{2}x$

14. On heating the amorphous substance it gets changed to the crystalline form at some temperature. This is due to the process called crystallization. As on heating at some temperature it may become crystalline since slow heating and cooling over a longer period of time makes these changes.

The given D.E. is

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

$\Rightarrow \frac{dx}{dy}-\frac{1}{y}.x=3y$ Which is form $\frac{dx}{dy}+Px=Q$

So, $P=-\frac{1}{y}\&Q=3y$

$\therefore I.F=e^{{\displaystyle \int Pdy}}=e^{{\displaystyle \int -\frac{1}{y}dy}}=e^{-log\left|y\right|}=e^{log{y}^{-1}}=y^{-1}=\frac{1}{y}$

Thus the solution is of the form.

$\begin{array}{l}x\times \frac{1}{y}={\displaystyle \int 3y.\frac{1}{y}dy+c}\\ =\frac{x}{y}={\displaystyle \int 3dy+c}\\ =\frac{x}{y}=3y+c\\ =x=3y^2+cy\end{array}$

Candidates often find it difficult to select college/ university during the admission season. The way to make a decision is college comparison in a fair manner. Motherhood University BPharm comparison can also be done based on certain parameters. Candidates can check below a few parameters that can be considered for the same:

• Fees
• Placements
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No, SRM University Andhra Pradesh does not offer direct admissions. Those who wish to take admission must follow the admission process designed by the university. As per the admission process, first, students who fulfil the eligibility requirements must apply for the course by visiting the official website of the university. The fee to apply is INR 1,400.

Once applied, students are required to wait for the merit list. Those who get shortlisted for the personal interview process must appear for the PI session on the scheduled date and time. Students who pass the round must complete the admission confirmation process.