Answer to Question #92900 in Differential Equations for Subhasis Padhy

Question

Solve.
d²y/dx² - 4x dy/dx + (4x²-1)y = -3exp(x²)sin(2x)

Answer 1

$y^{\prime\prime}-4xy^{\prime}+(4x^2-1)y=-3e^{x²}sin(2x)\\$ consider:$y^{\prime\prime}-4xy^{\prime}+(4x^2-1)y=0\\ y(x)=e^{x^2}\cdot z(x)\\ (2e^{x^2}z+4x^2e^{x^2}z+4xe^{x^2}z^{\prime}+e^{x^2}z^{\prime\prime})-4x(2xe^{x^2}z+e^{x^2}z^{\prime})+(4x^2-1)e^{x^2}z=0\\ z^{\prime\prime}+z=0\\ z=c_1\cos(x)+c_2\sin(x)\\ y(x)=c_1e^{x^2}\cos(x)+c_2e^{x^2}\sin(x)$

check the particular solution:$y_1=e^{x^2}\sin(2x)\\ y^\prime = 2\,x{{\rm e}^{{x}^{2}}}\sin \left( 2\,x \right) +2\,{{\rm e}^{{x}^{2}} }\cos \left( 2\,x \right)\\ y^{\prime\prime} = -2\,{{\rm e}^{{x}^{2}}}\sin \left( 2\,x \right) +4\,{x}^{2}{{\rm e}^{{ x}^{2}}}\sin \left( 2\,x \right) +8\,x{{\rm e}^{{x}^{2}}}\cos \left( 2 \,x \right) \\ y^{\prime\prime}-4xy^{\prime}+(4x^2-1)y =\\ -2\,{{\rm e}^{{x}^{2}}}\sin \left( 2\,x \right) +4\,{x}^{2}{{\rm e}^{{ x}^{2}}}\sin \left( 2\,x \right) +8\,x{{\rm e}^{{x}^{2}}}\cos \left( 2 \,x \right) -4x(2\,x{{\rm e}^{{x}^{2}}}\sin \left( 2\,x \right) +2\,{{\rm e}^{{x}^{2}} }\cos \left( 2\,x \right) )+(4x^2-1)(e^{x^2}\sin(2x))=-3e^{x^2}\sin(2x)$

then solution is:$y(x)=c_1e^{x^2}\cos(x)+c_2e^{x^2}\sin(x)+e^{x^2}\sin(2x)$