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Question: The differential equation $\{xy^3(1+\cos x)-y\}dx+xdy=0$ represents the curve $\frac{x^2}{ay^2}=\fra...

The differential equation {xy3(1+cos⁡x)−y}dx+xdy=0\{xy^3(1+\cos x)-y\}dx+xdy=0 represents the curve x2ay2=x3b+x2sin⁡x+cxcos⁡x−dsin⁡x+k\frac{x^2}{ay^2}=\frac{x^3}{b}+x^2\sin x+cx\cos x-d\sin x+k. Then a+b+c+da+b+c+d is equal to:

A

9

B

7

C

5

D

14−823\frac{14-8\sqrt{2}}{3}

Answer

9

Explanation

Solution

The given differential equation is: {xy3(1+cos⁡x)−y}dx+xdy=0\{xy^3(1+\cos x)-y\}dx+xdy=0

Rearrange the terms to isolate xdyxdy and ydxydx: xdy=−{xy3(1+cos⁡x)−y}dxxdy = -\{xy^3(1+\cos x)-y\}dx xdy=−xy3(1+cos⁡x)dx+ydxxdy = -xy^3(1+\cos x)dx + ydx xdy−ydx=−xy3(1+cos⁡x)dxxdy - ydx = -xy^3(1+\cos x)dx

This equation resembles the differential of yx\frac{y}{x} or −xy-\frac{x}{y}. Let's divide by x2x^2: xdy−ydxx2=−xy3(1+cos⁡x)x2dx\frac{xdy - ydx}{x^2} = -\frac{xy^3(1+\cos x)}{x^2}dx d(yx)=−y3x(1+cos⁡x)dxd\left(\frac{y}{x}\right) = -\frac{y^3}{x}(1+\cos x)dx

Let v=yxv = \frac{y}{x}. Then y=vxy=vx. Substitute this into the equation: dv=−(vx)3x(1+cos⁡x)dxdv = -\frac{(vx)^3}{x}(1+\cos x)dx dv=−v3x3x(1+cos⁡x)dxdv = -\frac{v^3 x^3}{x}(1+\cos x)dx dv=−v3x2(1+cos⁡x)dxdv = -v^3 x^2(1+\cos x)dx

This is a variable separable differential equation. Separate the variables vv and xx: dvv3=−x2(1+cos⁡x)dx\frac{dv}{v^3} = -x^2(1+\cos x)dx

Now, integrate both sides: ∫v−3dv=−∫(x2+x2cos⁡x)dx\int v^{-3} dv = -\int (x^2 + x^2\cos x)dx v−2−2=−[∫x2dx+∫x2cos⁡xdx]+C1\frac{v^{-2}}{-2} = -\left[\int x^2 dx + \int x^2\cos x dx\right] + C_1 −12v2=−[x33+∫x2cos⁡xdx]+C1-\frac{1}{2v^2} = -\left[\frac{x^3}{3} + \int x^2\cos x dx\right] + C_1 12v2=x33+∫x2cos⁡xdx−C1\frac{1}{2v^2} = \frac{x^3}{3} + \int x^2\cos x dx - C_1 Let C=−C1C = -C_1 (constant of integration). 12v2=x33+∫x2cos⁡xdx+C\frac{1}{2v^2} = \frac{x^3}{3} + \int x^2\cos x dx + C

Next, we need to evaluate the integral ∫x2cos⁡xdx\int x^2\cos x dx using integration by parts. The formula for integration by parts is ∫udv=uv−∫vdu\int u dv = uv - \int v du.

First, for ∫x2cos⁡xdx\int x^2\cos x dx: Let u=x2u=x^2 and dv=cos⁡xdxdv=\cos x dx. Then du=2xdxdu=2x dx and v=sin⁡xv=\sin x. ∫x2cos⁡xdx=x2sin⁡x−∫(sin⁡x)(2x)dx=x2sin⁡x−2∫xsin⁡xdx\int x^2\cos x dx = x^2\sin x - \int (\sin x)(2x) dx = x^2\sin x - 2\int x\sin x dx

Now, we need to evaluate ∫xsin⁡xdx\int x\sin x dx using integration by parts again: Let u=xu=x and dv=sin⁡xdxdv=\sin x dx. Then du=dxdu=dx and v=−cos⁡xv=-\cos x. ∫xsin⁡xdx=x(−cos⁡x)−∫(−cos⁡x)dx=−xcos⁡x+∫cos⁡xdx=−xcos⁡x+sin⁡x\int x\sin x dx = x(-\cos x) - \int (-\cos x) dx = -x\cos x + \int \cos x dx = -x\cos x + \sin x

Substitute this back into the expression for ∫x2cos⁡xdx\int x^2\cos x dx: ∫x2cos⁡xdx=x2sin⁡x−2(−xcos⁡x+sin⁡x)\int x^2\cos x dx = x^2\sin x - 2(-x\cos x + \sin x) ∫x2cos⁡xdx=x2sin⁡x+2xcos⁡x−2sin⁡x\int x^2\cos x dx = x^2\sin x + 2x\cos x - 2\sin x

Substitute this result back into the main solution equation: 12v2=x33+(x2sin⁡x+2xcos⁡x−2sin⁡x)+C\frac{1}{2v^2} = \frac{x^3}{3} + (x^2\sin x + 2x\cos x - 2\sin x) + C

Finally, substitute v=yxv = \frac{y}{x} back into the equation: 12(y/x)2=x33+x2sin⁡x+2xcos⁡x−2sin⁡x+C\frac{1}{2(y/x)^2} = \frac{x^3}{3} + x^2\sin x + 2x\cos x - 2\sin x + C x22y2=x33+x2sin⁡x+2xcos⁡x−2sin⁡x+C\frac{x^2}{2y^2} = \frac{x^3}{3} + x^2\sin x + 2x\cos x - 2\sin x + C

Now, compare this solution with the given form of the curve: x2ay2=x3b+x2sin⁡x+cxcos⁡x−dsin⁡x+k\frac{x^2}{ay^2}=\frac{x^3}{b}+x^2\sin x+cx\cos x-d\sin x+k

By comparing the coefficients of corresponding terms:

  1. Coefficient of x2y2\frac{x^2}{y^2}: 1a=12  ⟹  a=2\frac{1}{a} = \frac{1}{2} \implies a=2
  2. Coefficient of x31\frac{x^3}{1}: 1b=13  ⟹  b=3\frac{1}{b} = \frac{1}{3} \implies b=3
  3. Coefficient of x2sin⁡xx^2\sin x: 1=11=1 (matches)
  4. Coefficient of xcos⁡xx\cos x: c=2c=2
  5. Coefficient of sin⁡x\sin x: −d=−2  ⟹  d=2-d=-2 \implies d=2
  6. Constant term: k=Ck=C

We need to find the value of a+b+c+da+b+c+d: a+b+c+d=2+3+2+2=9a+b+c+d = 2+3+2+2 = 9