Question

solue the following differential equations"
dy/dx=cos^3(x).sin^4(x)+x√2x+1​

Answer 1

Question:-

$\\=> dy/dx=cos^3(x).sin^4(x)+x \sqrt{2x+1 }$

Solution :-

Let ,

Sin x = t

cos x dx = dt

And ,

2x +1 = u => x = u-1/2

2dx = du

Now,

$=> dy = (1-t^2 ) t^4 dt + \frac{1}{2} \sqrt{u} .\frac{u-1}{2} du\\\\\\=> \int dy = \int (t^4 -t^6 )dt + \frac{1}{4} \int (u^{3/2} - u^{1/2} )du\\\\\\=> y = \frac{t^5}{5} - \frac{t^7}{7} + \frac{1}{4} (\frac{2}{5} u^{5/2} - \frac{2}{5} u^{3/2} )$

Now, substitute  value of 't' and 'u 'in the above equation :

$\\=> y = \frac{ sin^5 x }{5} - \frac{sin^7 x}{7} + \frac{1}{10} (2x+1)^{5/2}$

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Answer 2

Question:-

$\\=> dy/dx=cos^3(x).sin^4(x)+x \sqrt{2x+1 }$

Solution :-

Let ,

Sin x = t

cos x dx = dt

And ,

2x +1 = u => x = u-1/2

2dx = du

Now,

$=> dy = (1-t^2 ) t^4 dt + \frac{1}{2} \sqrt{u} .\frac{u-1}{2} du\\\\\\=> \int dy = \int (t^4 -t^6 )dt + \frac{1}{4} \int (u^{3/2} - u^{1/2} )du\\\\\\=> y = \frac{t^5}{5} - \frac{t^7}{7} + \frac{1}{4} (\frac{2}{5} u^{5/2} - \frac{2}{5} u^{3/2} )$

Now, substitute  value of 't' and 'u 'in the above equation :

$\\=> y = \frac{ sin^5 x }{5} - \frac{sin^7 x}{7} + \frac{1}{10} (2x+1)^{5/2}$

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