Question

If $xdy - y \log_e ydx + 2yx^3 \sin xdx + yx^4 \cos xdx = 0$, then solution is

• A. $y = e^{x}(c-x^2 \tan x)$
• B. $y = e^{x^2}(c-x \sin x)$
• C. $y = e^{x}(c+x^2 \sin x)$
• D. $y = e^{x^2}(c+x \sin x)$

Answer 1

None of the given options are correct. The correct solution is $\ln y = cx - x^3\sin x$, or $y=e^{cx-x^3\sin x}$.

Answer 2

Here's how to solve the differential equation and why none of the provided options match the correct solution:

1. Rewrite the Equation:

   The given differential equation is:

   $$x\,dy - y\ln y\,dx + 2yx^3\sin x\,dx + yx^4\cos x\,dx=0$$

   Group the $dx$ terms:

   $$x\,dy + y\Bigl(2x^3\sin x + x^4\cos x -\ln y\Bigr)dx=0$$

2. Divide and Substitute:

   Divide the entire equation by $xy$ (assuming $x \ne 0$ and $y > 0$):

   $$\frac{dy}{y} + \Bigl[2x^2\sin x + x^3\cos x -\frac{\ln y}{x}\Bigr]dx=0$$

   Substitute $u = \ln y$, so $\frac{dy}{y} = du$. The equation becomes:

   $$du + \Bigl[2x^2\sin x + x^3\cos x -\frac{u}{x}\Bigr]dx=0$$

   Rearrange to get a linear ODE:

   $$\frac{du}{dx} - \frac{1}{x}\,u = -2x^2\sin x - x^3\cos x$$

3. Solve the Linear ODE:

   The integrating factor is:

   $$\mu(x)=e^{-\int\frac{1}{x}dx} = e^{-\ln|x|} = \frac{1}{x}$$

   Multiply the ODE by $\mu(x) = \frac{1}{x}$:

   $$\frac{1}{x}\frac{du}{dx} - \frac{1}{x^2}\,u = -2x\sin x - x^2\cos x$$

   The left side is the derivative of $\frac{u}{x}$:

   $$\frac{d}{dx}\Bigl(\frac{u}{x}\Bigr)= -2x\sin x - x^2\cos x$$

   Integrate both sides:

   $$\frac{u}{x}=-\int\Bigl[2x\sin x+x^2\cos x\Bigr]dx$$

   Since $\frac{d}{dx}(x^2\sin x)=2x\sin x+x^2\cos x$, the integral is:

   $$\int\Bigl[2x\sin x+x^2\cos x\Bigr]dx=x^2\sin x$$

   So,

   $$\frac{u}{x}=-x^2\sin x + c,\quad \text{or}\quad u=-x^3\sin x + cx$$

4. Back-Substitute:

   Substitute back $u = \ln y$:

   $$\ln y = cx - x^3\sin x,\quad \text{or}\quad y = e^{cx-x^3\sin x}$$

5. Comparison with Options:

   None of the provided options match the derived solution. The correct general solution is of the form $y = e^{cx - x^3 \sin x}$.