$$x^2\frac{\text{d}^2y}{\text{d}x^2}+x\frac{\text{d}y}{\text{d}x}+y=2sin(\ln x) ,x>0$$ answer is : $$c_1 sin(\ln x)+c_2cos(\ln x)-\ln x .cos(\ln x)$$ How do you solve this?How would this be solved? I got it so the equation in standard form would be y"+y'/x+y/x^2=2sin(lnx)
Here is a good start:
Solve the homogeneous part first: $$x^2 \frac{d^2 y}{dx^2}+x\frac{dy}{dx}+y=0$$ Use the ansatz $y=x^{\lambda}$ to obtain: $$x^2 \lambda(\lambda-1)x^{\lambda-2}+x\lambda x^{\lambda-1}+x^{\lambda}=0$$ $${\lambda}^2 x^{\lambda}+x^{\lambda}=0$$ $$\lambda^2+1=0$$ Solving for $\lambda$ gives the solution to the homogeneous part $y_c$: $$y_c=c_1 x^{i}+c_2 x^{-i}$$ We can use the fact that $x^i=e^{i\ln x}$ then apply Euler's identity. Then, redefining constants, we obtain the general solution to the homogeneous part: $$y_c=k_1\cos(\ln x)+k_2 \sin(\ln x)$$ Notice that this looks similar to the solution you were given so far. Now, to solve for the particular solution $y_p$, use Variation of Parameters and use the fact that $y=y_c+y_p$.
Feel free to tell us your progress on solving this problem.