Is it necessary to practice all NCERT exercises in Differential Equations to get good marks in CBSE Board Exams?

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

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)

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

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