Question

solve dy/dx + x sin 2y = x^3 cos^2 y​

Answer 1

Answer:

Step-by-step explanation:

The given equation becomes [(sec(y))^2].y' + 2x(tan y) = x^3, which after a substitution of tan(y) = u, becomes a linear differential equation: u' + 2x.u = x^3, which has an integrating factor of e^[Int(2x.dx)] = e^(x^2) and the general solution is

u.e^(x^2) = (1/2)Int[(x^2).e^(x^2)].2x.dx + c. The integral on the right is solved using a substitution x^2 = t.

Answer 2

Given:

          $\frac{d y}{d x}+x \sin 2 y=x^{3} \cos ^{2} y$

Solution:

         $\frac{d y}{d x}+x \sin 2 y=x^{3} \cos ^{2} y$

Dividing both sides by $cos ^{2} \mathrm{x}$

       $\sec ^{2} x \frac{d y}{d x}+x \sec ^{2} x \sin 2 y=x^{3}$

      $\sec ^{2} x \frac{d y}{d x}+2 x \tan y=x^{3} \ldots \ldots \ldots \ldots(1)$

Put $tan y=v$ and differentiate $w.r.t. x$

          $\sec ^{2} x \frac{d y}{d x}=\frac{d v}{d x}$

Hence, from ( 1 ), we get

                $\frac{d v}{d x}+2 v \cdot x=x^{3}$

          $P=2 x \\And Q=x^{3}$

$\therefore \int P d x=\int 2 x d x=x^{2}$

$I.F. =e^{\int P d x}=e^{\int 2 x d x}=e^{x^{2}}$

$\there The solution is v \ e^{x^{2}}=\int e^{x^{2}} x^{3} d x+c$

To find the integral put

                  $x^{2}=t, x d x=\frac{a t}{2}$

$\therefore I=\int e^{t} \cdot t \cdot \frac{d t}{2}=\frac{1}{2}\left[t e^{t}-\int e^{t} . d t\right] \ldots \ldots \ldots[ By \ parts ]$

$\therefore I=\frac{1}{2}\left[t e^{t}-e^{t}\right]=\frac{1}{2} e^{t}(t-1)=\frac{1}{2} e^{x^{2}}\left(x^{2}-1\right)$

$\there The solution is v \ e^{x^{2}}=\frac{1}{2} e^{x^{2}}\left(x^{2}-1\right)+c$

  $\therefore \tan y \ e^{x^{2}}=\frac{1}{2} e^{x^{2}}\left(x^{2}-1\right)+c$