Answer:-
y ′ ′ − 2 y ′ + y = e x arcsin x The auxiliary equation is m 2 − 2 m + 1 = 0 ( m − 1 ) 2 = 0 m = 1 twice y = e x ( C 1 + x C 2 ) Wronskian ( e x , x e x ) = e 2 x Let C 1 and C 2 be functions K 1 and K 2 . K 1 = − ∫ x e x × e x arcsin x e 2 x d x = − ∫ x arcsin x d x and K 2 = ∫ e x × e x arcsin x e 2 x d x = ∫ arcsin x d x For K 1 , Let x = sin u ⟹ d ( sin u ) = cos u d u K 1 = − ∫ x arcsin x d x = − ∫ u sin u cos u d u = − 1 2 ∫ u sin 2 u d u = 1 2 ∫ u d ( cos 2 u 2 ) = u cos 2 u 4 − 1 4 ∫ cos 2 u d u = u cos 2 u 4 − sin 2 u 8 + C = arcsin x cos ( 2 arcsin x ) 4 − sin ( 2 arcsin x ) 8 + C 1 For K 2 Let x = sin u ⟹ d ( sin u ) = cos u d u K 2 = ∫ arcsin x d x = ∫ u cos u d u = u sin u − ∫ sin u d u = u sin u + cos u + C = x arcsin x + cos ( arcsin x ) + C 2 But y = e x K 1 + x e x K 2 ⟹ y = e x ( arcsin x cos ( 2 arcsin x ) 4 − sin ( 2 arcsin x ) 8 + C 1 ) + x e x ( x arcsin x + cos ( arcsin x ) + C 2 ) = C 1 e x + C 2 x e x + e x arcsin x cos ( 2 arcsin x ) 4 − e x sin ( 2 arcsin x ) 8 + x 2 e x arcsin x + x e x cos ( arcsin x ) Note that the following holds cos ( 2 arcsin ( x ) ) = 1 − 2 x 2 sin ( 2 arcsin ( x ) ) = 2 x 1 − x 2 cos ( arcsin ( x ) ) = 1 − x 2 Substituting the above in y y = C 1 e x + C 2 x e x + e x arcsin x ( 1 − 2 x 2 ) 4 − 2 x e x 1 − x 2 8 + x 2 e x arcsin x + x e x 1 − x 2 = C 1 e x + C 2 x e x + e x arcsin x 4 − x 2 e x arcsin x 2 − x e x 1 − x 2 4 + x 2 e x arcsin x + x e x 1 − x 2 = C 1 e x + C 2 x e x + e x arcsin x 4 + x 2 e x arcsin x 2 + 3 x e x 1 − x 2 4 ⟹ y = C 1 e x + C 2 x e x + 3 x e x 1 − x 2 4 + x 2 e x arcsin x 2 + e x arcsin x 4 \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} y ′′ − 2 y ′ + y = e x arcsin x The auxiliary equation is m 2 − 2 m + 1 = 0 ( m − 1 ) 2 = 0 m = 1 twice y = e x ( C 1 + x C 2 ) Wronskian ( e x , x e x ) = e 2 x Let C 1 and C 2 be functions K 1 and K 2 . K 1 = − ∫ e 2 x x e x × e x arcsin x d x = − ∫ x arcsin x d x and K 2 = ∫ e 2 x e x × e x arcsin x d x = ∫ arcsin x d x For K 1 , Let x = sin u ⟹ d ( sin u ) = cos u d u K 1 = − ∫ x arcsin x d x = − ∫ u sin u cos u d u = − 2 1 ∫ u sin 2 u d u = 2 1 ∫ u d ( 2 cos 2 u ) = 4 u cos 2 u − 4 1 ∫ cos 2 u d u = 4 u cos 2 u − 8 sin 2 u + C = 4 arcsin x cos ( 2 arcsin x ) − 8 sin ( 2 arcsin x ) + C 1 For K 2 Let x = sin u ⟹ d ( sin u ) = cos u d u K 2 = ∫ arcsin x d x = ∫ u cos u d u = u sin u − ∫ sin u d u = u sin u + cos u + C = x arcsin x + cos ( arcsin x ) + C 2 But y = e x K 1 + x e x K 2 ⟹ y = e x ( 4 arcsin x cos ( 2 arcsin x ) − 8 sin ( 2 arcsin x ) + C 1 ) + x e x ( x arcsin x + cos ( arcsin x ) + C 2 ) = C 1 e x + C 2 x e x + 4 e x arcsin x cos ( 2 arcsin x ) − 8 e x sin ( 2 arcsin x ) + x 2 e x arcsin x + x e x cos ( arcsin x ) Note that the following holds cos ( 2 arcsin ( x )) = 1 − 2 x 2 sin ( 2 arcsin ( x )) = 2 x 1 − x 2 cos ( arcsin ( x )) = 1 − x 2 Substituting the above in y y = C 1 e x + C 2 x e x + 4 e x arcsin x ( 1 − 2 x 2 ) − 8 2 x e x 1 − x 2 + x 2 e x arcsin x + x e x 1 − x 2 = C 1 e x + C 2 x e x + 4 e x arcsin x − 2 x 2 e x arcsin x − 4 x e x 1 − x 2 + x 2 e x arcsin x + x e x 1 − x 2 = C 1 e x + C 2 x e x + 4 e x arcsin x + 2 x 2 e x arcsin x + 4 3 x e x 1 − x 2 ⟹ y = C 1 e x + C 2 x e x + 4 3 x e x 1 − x 2 + 2 x 2 e x arcsin x + 4 e x arcsin x
#333702 on Dec 2022
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