Answer to Question #315652 in Differential Equations for Sheshank

Question #315652

2. Solve by variation of parameters:



(D ^ 2 - 2D + 1) * y = x ^ (3/2) * e ^ x


1
Expert's answer
2022-03-23T14:39:24-0400

Homogeneous  equation:(D22D+1)y=0λ22λ+1=0λ1,2=1y=C1ex+C2xexu1(x)=ex,u2(x)=xexy(x)=A(x)u1(x)+B(x)u2(x)[u1(x)u2(x)u1(x)u2(x)][A(x)B(x)]=[0f(x)][exxexexex+xex][A(x)B(x)]=[0x3/2ex][A(x)B(x)]=[exxexexex+xex]1[0x3/2ex]==e2x[ex(1+x)xexexex][0x3/2ex]=[x5/2x3/2]A(x)=27x7/2,B=25x5/2y(x)=27x7/2ex+25x7/2ex=435x7/2exparticular  solutiony(x)=435x7/2ex+C1ex+C2xexgeneral  solutionHomogeneous\,\,equation:\\\left( D^2-2D+1 \right) y=0\\\lambda ^2-2\lambda +1=0\Rightarrow \lambda _{1,2}=1\Rightarrow \\\Rightarrow y=C_1e^x+C_2xe^x\\u_1\left( x \right) =e^x,u_2\left( x \right) =xe^x\\y\left( x \right) =A\left( x \right) u_1\left( x \right) +B\left( x \right) u_2\left( x \right) \\\left[ \begin{matrix} u_1\left( x \right)& u_2\left( x \right)\\ u_1'\left( x \right)& u_2'\left( x \right)\\\end{matrix} \right] \left[ \begin{array}{c} A'\left( x \right)\\ B'\left( x \right)\\\end{array} \right] =\left[ \begin{array}{c} 0\\ f\left( x \right)\\\end{array} \right] \\\left[ \begin{matrix} e^x& xe^x\\ e^x& e^x+xe^x\\\end{matrix} \right] \left[ \begin{array}{c} A'\left( x \right)\\ B'\left( x \right)\\\end{array} \right] =\left[ \begin{array}{c} 0\\ x^{3/2}e^x\\\end{array} \right] \Rightarrow \\\Rightarrow \left[ \begin{array}{c} A'\left( x \right)\\ B'\left( x \right)\\\end{array} \right] =\left[ \begin{matrix} e^x& xe^x\\ e^x& e^x+xe^x\\\end{matrix} \right] ^{-1}\left[ \begin{array}{c} 0\\ x^{3/2}e^x\\\end{array} \right] =\\=e^{-2x}\left[ \begin{matrix} e^x\left( 1+x \right)& -xe^x\\ -e^x& e^x\\\end{matrix} \right] \left[ \begin{array}{c} 0\\ x^{3/2}e^x\\\end{array} \right] =\left[ \begin{array}{c} -x^{5/2}\\ x^{3/2}\\\end{array} \right] \\A\left( x \right) =-\frac{2}{7}x^{7/2},B=\frac{2}{5}x^{5/2}\\y\left( x \right) =-\frac{2}{7}x^{7/2}e^x+\frac{2}{5}x^{7/2}e^x=-\frac{4}{35}x^{7/2}e^x-particular\,\,solution\\y\left( x \right) =-\frac{4}{35}x^{7/2}e^x+C_1e^x+C_2xe^x-general\,\,solution


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