Do NCERT Class 12 Differential Equations Solutions help in NDA preparation?

  $\frac{dy}{dx}-\left(\frac{sin2x}{1+co{s}^{2}x}\right)y=\frac{sinx}{1+co{s}^{2}x}$

IF = $e^{-{\displaystyle \int \frac{sin2xdx}{1+co{s}^{2}x}}}$  

$=e^{ln\left(1+co{s}^{2}x\right)}=\left(1+co{s}^{2}x\right)$        

So, y(1 + cos² x) = ${\displaystyle \int \frac{sinx}{\left(1+co{s}^{2}x\right)}\cdot \left(1+co{s}^{2}x\right)dx}$  

y(1 + cos² x) = – cos x + c

$?$      y(0) = 0

0 = – 1 + c

-> c = 1

$y=\frac{1-cosx}{1+co{s}^{2}x}$   

Now, $y\left(\frac{\pi }{2}\right)=1$  

$\frac{dy}{dx}=\frac{\left(x+1\right)\left(x^2-x+1\right)+\sqrt[]{\left(1-x\right)\left(1+x\right)}}{\left(x-1\right)\left(x+1\right)}$

$\frac{dy}{dx}=\frac{x\left(x-1\right)+1}{\left(x-1\right)}+\sqrt[]{\frac{\left(1-x\right)\left(1+x\right)}{{\left(x-1\right)}^2{\left(x+1\right)}^2}}$

$\frac{dy}{dx}=x+\frac{1}{x-1}+\frac{1}{\sqrt[]{\left(1-x\right)\left(1+x\right)}}$

$dy=xd+\frac{1}{\left(x-1\right)}dx+\frac{dx}{\sqrt[]{1-x^2}}$

$y=\frac{x^2}{2}+ln\left|x-1\right|+si{n}^{-1}x+c$

at x = 0, y = 2 2 = c

$y=\frac{x^2}{2}+ln\left|x-1\right|+si{n}^{-1}x+2$

$y\left(\frac{1}{2}\right)=\frac{17}{8}+\frac{\pi }{6}-ln2$

5f(x) + 4f $\left(\frac{1}{x}\right)$  = x² – 4           ...(1)

Replace x by  $\frac{1}{x}$

5f  $\left(\frac{1}{x}\right)$  + 4f(x) = $\frac{1}{x^2}$  – 4   ...(2)

5 × equation (1) – 4 × equation (2)

$9f\left(x\right)=5x^{2-}\frac{4}{x^2}-4$            

$y=9f\left(x\right)\cdot x^2=\frac{5x^4-4-4x^2}{x^2}{x}^2$            

y = 5x⁴ – 4 – 4x²

y = 20x³ – 8x > 0

4x(5x² – 2) > 0

    $x\in \left(-\sqrt[]{\frac{2}{5}},0\right)\cup \left(\sqrt[]{\frac{2}{5}},\infty \right)$

(t + 1)dx = (2x + (t + 1)³)dt

$\frac{dx}{dt}-\frac{2x}{t+1}={\left(t+1\right)}^2$

I.F. $=e^{{\displaystyle \int -\frac{2}{t+1}dt}}=\frac{1}{{\left(t+1\right)}^2}$  

Solution is

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

x = (t + c) (t + 1)²

$?$ x (0) = 2 then c = 2

x = (t + 2) (t + 1)²

 x (1) = 12

$\frac{dp}{dt}=0.5p-450andP\left(0\right)=850$

$\Rightarrow \frac{dp}{P-900}=0.5dt$

${\displaystyle \underset{850}{\overset{0}{\int }}\frac{dp}{P-900}={\displaystyle \underset{0}{\overset{T}{\int }}0.5dt}}$

$\Rightarrow ln\left(P-900\right){|}_{805}^0=0.5T$

$\frac{T}{2}=ln\left|\frac{900}{50}\right|=ln18$

T = 2 ln 18