Answer to Question #86129 in Electric Circuits for Ajay

Question

Solve the differential equation: d^2y/dx^2+dy/dx+y=3sin3x

Answer 1

Using Lagrange method (variation of constants),

$$A(x) = -\int \frac{1}{W} y_2(x) 3 \sin(3x)dx = -2\sqrt{3} \int e^{\frac{x}{2}} \sin{\frac{\sqrt{3}x}{2}}\sin{3x}dx$$

$$B(x) = \int \frac{1}{W} y_1(x) 3 \sin(3x)dx = 2\sqrt{3} \int e^{\frac{x}{2}} \cos{\frac{\sqrt{3}x}{2}}\sin{3x}dx$$

$$A(x) = \sqrt{3}e^{x/2}\left(\frac{\left(\sqrt{3}-6\right) \sin \left(\frac{1}{2} \left(\sqrt{3}-6\right) x\right)+\cos \left(\frac{1}{2} \left(\sqrt{3}-6\right) x\right)}{6 \sqrt{3}-20}+\frac{\left(\sqrt{3}+6\right) \sin \left(\frac{1}{2} \left(\sqrt{3}+6\right) x\right)+\cos \left(\frac{1}{2} \left(\sqrt{3}+6\right) x\right)}{6 \sqrt{3}+20}\right.$$

$$B(x) = \sqrt{3} e^{x/2} \left(\frac{\sin \left(\frac{1}{2} \left(\sqrt{3}+6\right) x\right)-\left(\sqrt{3}+6\right) \cos \left(\frac{1}{2} \left(\sqrt{3}+6\right) x\right)}{6 \sqrt{3}+20}-\frac{\sin \left(3 x-\frac{\sqrt{3} x}{2}\right)+\left(\sqrt{3}-6\right) \cos \left(\frac{1}{2} \left(\sqrt{3}-6\right) x\right)}{6 \sqrt{3}-20}\right)$$

Substituting A(x) and B(x) to the general solution of the homogeneous eq. we get general solution of the nonhomogeneous eq.:

$$y(x)= c_1e^{-\frac{x}{2}} \sin \left(\frac{\sqrt{3} x}{2}\right)+c_2 e^{-\frac{x}{2}} \cos \left(\frac{\sqrt{3} x}{2}\right)-\frac{3}{73} (8 \sin (3 x)+3 \cos (3 x))$$