Answer to Question #223344 in Differential Equations for Anuj

Question

Find the general solution of following

1) y''-2y'+y=e^xsin^-1(x)

Answer 1

$\displaystyle y'' - 2y' + y = e^x\arcsin{x}\\ \textsf{The auxiliary equation is}\\ m^2 - 2m + 1 = 0\\ (m - 1)^2 = 0\\ m = 1 \,\, \textsf{twice}\\ y = e^x(C_1 + xC_2)\\ \textsf{Wronskian}(e^x, xe^x) = e^{2x}\\ \textsf{Let} \,\, C_1 \,\,\textsf{and}\,\, C_2 \,\, \textsf{be}\,\, \textsf{functions}\\ K_1 \,\,\textsf{and}\,\, K_2.\\ K_1 = - \int \frac{xe^x \times e^{x}\arcsin{x}}{e^{2x}}\mathrm{d}x = -\int x\arcsin{x}\mathrm{d}x \\ \textsf{and} \,\, K_2 = \int \frac{e^x \times e^{x}\arcsin{x}}{e^{2x}}\mathrm{d}x = \int \arcsin{x}\mathrm{d}x \\ \textsf{For}\,\, K_1, \,\, \textsf{Let}\,\, x = \sin{u}\\ \implies \mathrm{d}(\sin{u}) = \cos{u}\,\,\mathrm{d}u\\ \begin{aligned} K_1 &= -\int x\arcsin{x}\mathrm{d}x = -\int u\sin{u}\cos{u}\mathrm{d}u \\&= -\frac{1}{2}\int u\sin{2u}\mathrm{d}u \\&= \frac{1}{2}\int u\mathrm{d}\left(\frac{\cos{2u}}{2}\right) \\&= \frac{u\cos{2u}}{4} - \frac{1}{4}\int \cos{2u}\mathrm{d}u \\&= \frac{u\cos{2u}}{4} - \frac{\sin{2u}}{8} + C \\&= \frac{\arcsin{x}\cos{\left(2\arcsin{x}\right)}}{4} - \frac{\sin\left(2\arcsin{x}\right)}{8} + C_1 \end{aligned}\\ \textsf{For}\,\, K_2 \,\, \textsf{Let}\,\, x = \sin{u}\\ \implies \mathrm{d}(\sin{u}) = \cos{u}\,\,\mathrm{d}u\\ \begin{aligned} K_2 &= \int \arcsin{x}\mathrm{d}x = \int u\cos{u}\mathrm{d}u \\&= u\sin{u} - \int \sin{u}\mathrm{d}u \\&= u\sin{u} + \cos{u} + C \\&= x\arcsin{x} + \cos\left(\arcsin{x}\right) + C_2 \end{aligned}\\ \textsf{But}\,\,\, y = e^x K_1 + xe^xK_2\\ \begin{aligned} \implies y &= e^x\left(\frac{\arcsin{x}\cos{\left(2\arcsin{x}\right)}}{4} -\frac{\sin\left(2\arcsin{x}\right)}{8} + C_1\right) \\&\hspace{2cm}+ xe^x\left(x\arcsin{x} + \cos\left(\arcsin{x}\right) + C_2\right) \\&= C_1e^x + C_2xe^x + \frac{e^x\arcsin{x}\cos{\left(2\arcsin{x}\right)}}{4} - \frac{e^x\sin\left(2\arcsin{x}\right)}{8} \\&\hspace{2cm}+ x^2e^x\arcsin{x} + xe^x\cos\left(\arcsin{x}\right) \end{aligned}\\ \textsf{Note that the following holds}\\ \cos(2\arcsin(x)) = 1 - 2x^2\\ \sin(2\arcsin(x)) = 2x\sqrt{1 - x^2}\\ \cos(\arcsin(x)) = \sqrt{1 - x^2}\\ \textsf{Substituting the above in}\,\,\, y\\ \begin{aligned} y &= C_1e^x + C_2xe^x + \frac{e^x\arcsin{x}(1 - 2x^2)}{4} - \frac{2xe^x\sqrt{1 - x^2}}{8} \\&\hspace{2cm}+ x^2e^x\arcsin{x} + xe^x\sqrt{1 - x^2} \\&= C_1e^x + C_2xe^x + \frac{e^x\arcsin{x}}{4} - \frac{x^2e^x\arcsin{x}}{2} - \frac{xe^x\sqrt{1 - x^2}}{4} \\&\hspace{2cm}+ x^2e^x\arcsin{x} + xe^x\sqrt{1 - x^2} \\&= C_1e^x + C_2xe^x + \frac{e^x\arcsin{x}}{4} + \frac{x^2e^x\arcsin{x}}{2} + \frac{3xe^x\sqrt{1 - x^2}}{4} \end{aligned}\\ \implies y = C_1e^x + C_2xe^x + \frac{3xe^x\sqrt{1 - x^2}}{4}+ \frac{x^2e^x\arcsin{x}}{2} + \frac{e^x\arcsin{x}}{4}$