Answer to Question #302579 in Differential Equations for haru

Question #302579

Find the general solution of the following differential equations using method of undetermined coefficients: , (ii) 2y''−5y' = x^2 +5e^−4x,


1
Expert's answer
2022-02-28T17:54:20-0500

Corresponding homogeneous differential equation


2y5y=02y''−5y' =0

Characteristic (auxiliary) equation


2r25r=02r^2-5r=0

r1=0,r2=5/2r_1=0,r_2=5/2

The general solution of the homogeneous differential equation is


yh=c1+c2e5x/2y_h=c_1+c_2e^{5x/2}

Find the particular solution of the non homogeneous differential equation


yp=Ax3+Bx2+Cx+De4xy_p=Ax^3+Bx^2+Cx+De^{-4x}

yp=3Ax2+2Bx+C4De4xy_p'=3Ax^2+2Bx+C-4De^{-4x}

yp=6Ax+2B+16De4xy_p''=6Ax+2B+16De^{-4x}

Substitute


6Ax+2B+16De4x6Ax+2B+16De^{-4x}

15Ax210Bx5C+20De4x-15Ax^2-10Bx-5C+20De^{-4x}

=x2+5e4x=x^2+5e^{-4x}

15A=1-15A=1

6A10B=06A-10B=0

2B5C=02B-5C=0

36D=536 D=5

A=1/15,B=1/25,C=2/125,D=5/36A=-1/15, B=-1/25, C=-2/125, D=5/36

yp=115x3125x22125x+536e4xy_p=-\dfrac{1}{15}x^3-\dfrac{1}{25}x^2-\dfrac{2}{125}x+\dfrac{5}{36}e^{-4x}

The general solution of the non homogeneous differential equation is


y=c1+c2e5x/2115x3125x22125x+536e4xy=c_1+c_2e^{5x/2}-\dfrac{1}{15}x^3-\dfrac{1}{25}x^2-\dfrac{2}{125}x+\dfrac{5}{36}e^{-4x}




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