Answer to Question #202277 in Differential Equations for king

Question

1.Solve Cos y cos px + sin y sin px = p

2.Solve

(D^3+2D^2+D)y= e^2x + x^2 + sin2x

Answer 1

1.

$$\cos y\cos (px)+\sin y\sin (px)=p$$

$$\cos(px-y)=p$$

$$px-y=\cos^{-1}(p)$$

$$y=px-\cos^{-1}(p), p=\dfrac{dy}{dx}$$

This is a Clairaut differential equation.

Differentiating with respect to $x,$ both sides 

$$p=p+x\dfrac{dp}{dx}+\dfrac{1}{\sqrt{1-p^2}}\cdot\dfrac{dp}{dx}$$

$$(x+\dfrac{1}{\sqrt{1-p^2}})\dfrac{dp}{dx}=0$$

Hence

$$x=-\dfrac{1}{\sqrt{1-p^2}} \text{ or }\dfrac{dp}{dx}=0=>p=c$$

$$x=-\dfrac{1}{\sqrt{1-p^2}}=>1-p^2=\dfrac{1}{x^2 }, \ x<0$$

$$p^2=\dfrac{x^2-1}{x^2}$$

$$p=\pm\dfrac{\sqrt{x^2-1}}{x}$$

Substitute

$$y=\pm(\sqrt{x^2-1}-\cos^{-1}(\dfrac{\sqrt{x^2-1}}{x})), -1\leq x<0$$

$$p=\dfrac{dy}{dx}=\dfrac{x}{\sqrt{x^2-1}}-\dfrac{1}{x\sqrt{x^2-1}}=\dfrac{\sqrt{x^2-1}}{x}$$

Putting $p=c$ in the equation we have

$$y=cx-\cos^{-1}(c)$$

2.

$$(D^3+2D^2+D)y=e^{2x}+x^2+\sin(2x)$$

$$y'=z, y''=z', y'''=z''$$

$$z''+2z'+z=e^{2x}+x^2+\sin(2x)$$

Homogeneous Equation

$$z''+2z'+z=0$$

The characteristic (auxiliary) equation

$$r^2+2r+1=0$$

$$r_1=r_2=-1$$

$$z_h=c_1xe^{-x}+c_2e^{-x}$$

$$z_p=Ae^{2x}+B\sin(2x)+C\cos(2x)$$

$$+(Mx^2+Nx+Q)$$

$$z_p'=2Ae^{2x}+2B\cos(2x)-2C\sin(2x)$$

$$+(2Mx+N)$$

$$z_p''=4Ae^{2x}-4B\sin(2x)-4C\cos(2x)+2M$$

Then

$$4Ae^{2x}-4B\sin(2x)-4C\cos(2x)+2M$$

$$+4Ae^{2x}+4B\cos(2x)-4C\sin(2x)+4Mx+2N$$

$$+Ae^{2x}+B\sin(2x)+C\cos(2x)+Mx^2+Nx+Q$$

$$=e^{2x}+x^2+\sin(2x)$$

$$A=\dfrac{1}{9}$$

$$-3B-4C=1$$

$$4B-3C=0$$

$$M=\dfrac{1}{2}$$

$$N=-2$$

$$Q=3$$

$$z_p=\dfrac{1}{9}e^{2x}-\dfrac{3}{25}\sin(2x)-\dfrac{4}{25}\cos(2x)$$

$$+\dfrac{1}{2}x^2-2x+3$$

Therefore

$$z(x)=c_1xe^{-x}+c_2e^{-x}+\dfrac{1}{9}e^{2x}$$

$$-\dfrac{3}{25}\sin(2x)-\dfrac{4}{25}\cos(2x)+\dfrac{1}{2}x^2-2x+3$$

$$\int xe^{-x}dx=-xe^{-x}+\int e^{-x}dx$$

$$=-xe^{-x}-e^{-x}+C$$

$$y(x)=c_5+c_3xe^{-x}+c_4e^{-x}+\dfrac{1}{18}e^{2x}$$

$$+\dfrac{3}{50}\cos(2x)-\dfrac{4}{50}\sin(2x)+\dfrac{1}{6}x^3-x^2+3x$$