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Yes, applications are open at Parle Tilak Vidyalaya Association's Sathaye College for various courses. The mode of application is online. Aspirants need to apply at the official website, i.e., enrollonline.co.in/Registration/Apply/SCA for admission to the various programme. Candidates can fill out the application form on the institute's official website.

The given D.E. is $y^1=e^{x}sinx$

$dy=e^{x}sinxdx$

Integrating both sides,

$\begin{array}{l}{\displaystyle \int dy={\displaystyle \int e^{x}sinxdx}}\\ \Rightarrow y=I+c\end{array}$

Where, $I={\displaystyle \int e^{x}sinxdx}$

$\begin{array}{l}=sinx{\displaystyle \int e^{x}dx-{\displaystyle \int \frac{d}{dx}sinx{\displaystyle \int e^{x}dx.dx}}}\\ =sinx.e^x-{\displaystyle \int cosx{e}^{x}dx}\\ =sinx{e}^x-\left\{cosx{\displaystyle \int e^{x}dx}-{\displaystyle \int \frac{d}{dx}\left(cosx\right).{\displaystyle \int I={\displaystyle \int e^{x}x}dx}}\right\}\\ =sinx.e^x-\left\{cosx{e}^x+{\displaystyle \int sinx{e}^{x}dx}\right\}\\ =sinx.e^x-cosx{e}^x-I\\ \Rightarrow I+I=e^x\left(sinx-cosx\right)\\ \Rightarrow I=\frac{e^x}{2}\left(sinx-cosx\right)+c\end{array}$

Hence, $y=\frac{e^x}{2}\left(sinx-cosx\right)+c$

When the curve passed point (0,0),

$\begin{array}{l}y=0,at,x=0\\ \Rightarrow 0=\frac{e^x}{2}\left(sin0-cos0\right)+c\\ \Rightarrow \frac{e^0}{2}\left(0-1\right)=c\\ \Rightarrow c=\frac{1}{2}\end{array}$

$\therefore$ The required equation of the curve is $y=\frac{e^x}{2}\left(sinx-cosx\right)+\frac{1}{2}$

$\begin{array}{l}\Rightarrow 2y=e^x\left(sinx-cosx\right)+1\\ \Rightarrow 2y-1=e^x\left(sinx-cosx\right)\end{array}$

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In order to get admission at Parle Tilak Vidyalaya Association's Sathaye College candidates must keep certain documents ready. Student can check the list of documents below:

• Scanned Class 10, Class 12 and graduation marksheet
• Leaving Certificate
• Aadhaar Card
• Passport-size photograph
• Signature

Note: The list of documents mentioned above are the general documents. For more details visit the official website.

108. Given, $y=A{e}^{mx}+B{e}^{nx}$ _______(1)

So, $\frac{dy}{dx}=Am{e}^{mx}+Bn{e}^{nx}$ _______(2)

$\therefore \frac{d^{2}y}{d{x}^2}=A{m}^2{e}^{mx}+B{n}^2{e}^{nx}$ _________(3)

So, L.H.S = $\frac{d^{2}y}{d{x}^2}-\left(m+n\right)\frac{dy}{dx}+mny$

$=A{m}^2{e}^{mx}+B{n}^2{e}^{nx}-\left(m+n\right)\left[Am{e}^{mx}+Bn{e}^{nx}\right]+mn\left[A{e}^{mx}+B{e}^{nx}\right]$

$=A{m}^2{e}^{mx}+B{n}^2{e}^{nx}-A{m}^2{e}^{mx}-Bmn{e}^{nx}-Amn{e}^{mx}-B{n}^2{e}^{nx}+Amn{e}^{mx}+Bmn{e}^{nx}$

= 0 = R.H.S.

Given,  $\frac{dy}{dx}=ytanx$

$\Rightarrow \frac{dy}{y}=tanxdx$

Integrating both sides we get,

$\begin{array}{l}{\displaystyle \int \frac{dy}{y}}={\displaystyle \int tanxdx}\\ \Rightarrow logy=log\left|secx\right|+logc\\ \Rightarrow logy=log\left|csecx\right|\\ \Rightarrow y=c_{1}secx\left(where,c_1=\pm c\right)\end{array}$

As,  $y=1,at,x=0$ we have,

$1=c_{1}sec\left(0\right)=c\Rightarrow c=1$

$\therefore$ The required particular solution is $y=secx$ .

Given, D.E. is

$\begin{array}{l}cos\frac{dy}{dx}=a\\ \Rightarrow \frac{dy}{dx}=co{s}^{-1}\left(a\right)\\ \Rightarrow dy=co{s}^{-1}\left(a\right)dx\end{array}$

Integrating both sides,

$\begin{array}{l}{\displaystyle \int dy={\displaystyle \int co{s}^{-1}\left(a\right)dx}}\\ \Rightarrow y=co{s}^{-1}\left(a\right)\times x+c\\ \Rightarrow y=xco{s}^{-1}\left(a\right)dx\end{array}$

Given,  $y=1,atx=0$

Then,  $1=0co{s}^{-1}\left(a\right)+c$

$\Rightarrow c=1$

$\therefore$ The required particular solution is

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Disclaimer: This information is sourced from official website and may vary.

The given D.E. is

$\begin{array}{l}x\left(x^2-1\right)\frac{dy}{dx}=1\\ \Rightarrow dy=\frac{dx}{x\left(x^2-1\right)}\end{array}$

Integrating both sides,

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

Let, $\frac{1}{x\left(x-1\right)\left(x+1\right)}=\frac{A}{x}+\frac{B}{x-1}+\frac{c}{x+1}$

$\begin{array}{l}1=A\left(x-1\right)\left(x+1\right)+B\left(x\right)\left(x+1\right)+C\left(x\right)\left(x-1\right)\\ =A\left(x^2-1\right)+B{x}^2+Bx+C{x}^2-Cx\\ =A{x}^2-A+B{x}^2+Bx+C{x}^2-Cx\\ =\left(A+B+C\right)x^2+\left(B-C\right)x-A\end{array}$

Comparing the coefficient,

$\begin{array}{l}-A=1\\ \Rightarrow A=-1------------(1)\\ \Rightarrow A+B+C=0---------(2)\\ \Rightarrow B-C=0\\ \Rightarrow B=C-------------(3)\end{array}$

Putting equation (1) & (2) in (1) we get,

$\begin{array}{l}-1+B+B=0\\ \Rightarrow -1+2B=0\\ \Rightarrow B=\frac{1}{2}=C\end{array}$

So, $\frac{1}{x\left(x-1\right)\left(x+1\right)}=-\frac{1}{x}+\frac{\frac{1}{2}}{x-1}+\frac{\frac{1}{2}}{x+1}$

$=-\frac{1}{x}+\frac{1}{2\left(x-1\right)}+\frac{1}{2\left(x+1\right)}$

Integrating becomes,

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

Given, $y=0whenx=2.$

Then, $0=\frac{1}{2}log\frac{2^2-1}{2^2}+c$

$\begin{array}{l}\Rightarrow 0=\frac{1}{2}log\frac{3}{4}+c\\ \Rightarrow c=-\frac{1}{2}log\frac{3}{4}\end{array}$

$\therefore$ The required particular solution is

$y=\frac{1}{2}log\frac{x^2-1}{x^2}-\frac{1}{2}log\frac{3}{4}$

107. Given $y=3cos\left(logx\right)+4sin\left(logx\right)$

So, $y_1=\frac{dy}{dx}=3\frac{d}{dx}cos\left(logx\right)+4\frac{d}{dx}sin\left(logx\right)$

$y_1=3\left[-sin\left(logx\right)\right]\frac{d}{dx}logx+4cos\left(logx\right)\frac{d}{dx}\left(logx\right)$

$y_1=\frac{-3sin\left(logx\right)}{x}+\frac{4cos\left(logx\right)}{x}$

$\Rightarrow x{y}_1=-3sin\left(logx\right)+4cos\left(logx\right)$ ______________(1)

Differentiating eqn (1) w r t ‘x’ we get,

$\frac{d}{dx}\left(x{y}_1\right)=-3\frac{d}{dx}sin\left(logx\right)+4\frac{d}{dx}cos\left(logx\right)$

$\Rightarrow x\frac{d{y}_1}{dx}+y_1\frac{dx}{dx}=-3cos\left(log\left(x\right)\right)\frac{d}{dx}logx+4\left[-sin\left(logx\right)\right]\frac{d}{dx}logx$

$\Rightarrow x{y}_2+y_1=\frac{-3cos\left(logx\right)}{x}-\frac{4sin\left(logx\right)}{x}$

$\Rightarrow x^2{y}_2+y_1=-\left[3cos\left(logx\right)+4sin\left(logx\right)\right]$

$\Rightarrow x^2{y}_2+y_1=-y$

$\Rightarrow x^2{y}_2+y_1+y=0$

The given D.E is $\left(x^3+x^2+x+1\right)\frac{dy}{dx}=2x^2+x$

$\begin{array}{l}\Rightarrow dy=\left(\frac{2x^2+x}{x^3+x^2+x+1}\right)dx\\ \Rightarrow dy=\frac{2x^2+x}{x^2\left(x+1\right)+\left(x+1\right)}dx=\frac{2x^2+x}{\left(x+1\right)\left(x^2+1\right)}dx\end{array}$

Integrating both sides we get,

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

Let, $\frac{2x^2+x}{\left(x+1\right)\left(x^2+1\right)}=\frac{A}{x+1}+\frac{Bx+c}{x^2+1}$

$\begin{array}{l}\Rightarrow 2x^2+2=A\left(x^2+1\right)+\left(Bx+c\right)\left(x+1\right)\\ =A{x}^2+A+B{x}^2+Bx+Cx+C\\ =\left(A+B\right)x^2+\left(B+C\right)x^2+\left(A+C\right)\end{array}$

Comparing the co-efficient we get,

$\begin{array}{l}A+B=2------(1)\\ B+C=1------(2)\\ A+C=0------(3)\end{array}$

Subtracting equation (1) – (2), we get

$\begin{array}{l}A+B-\left(B+C\right)=2-1\\ \to A-C=1\end{array}$

But from equation (3) $A=-C$ so, we get,

$\begin{array}{l}A\left(-C\right)-C=1\\ \to -2C=1\\ \to C=-\frac{1}{2}\\ \&A=-\left(-\frac{1}{2}\right)=\frac{1}{2}\end{array}$

And putting value of A in equation (1),

$\begin{array}{l}\frac{1}{2}+B=2\\ \Rightarrow B=2-\frac{1}{2}=\frac{4-1}{2}=\frac{3}{2}\end{array}$

Putting value of A,B and C in

$\begin{array}{l}\frac{2x^2+x}{\left(x+1\right)\left(x^2+1\right)}=\frac{\frac{1}{2}}{x+1}+\frac{\frac{3}{2}x-\frac{1}{2}}{x^2+1}\\ =\frac{1}{2\left(x+1\right)}+\frac{3}{2}\left(\frac{x}{x^2+1}\right)-\frac{1}{2}\left(\frac{1}{x^2+1}\right)\end{array}$