# Solve ( D 2 + 2 ) Y = E X Cos X + X 2 E 3 X - Applied Mathematics 2

Sum

Solve (D^2+2)y=e^xcosx+x^2e^(3x)

#### Solution

(D^2+2)y=e^xcosx+x^2e^(3x)
For complementary solution,

๐(๐ซ)=๐

therefore(D^2+2)=0
Roots are : D = √๐๐ ,−√๐๐

Roots of given diff. eqn are complex.
The complementary solution of given diff. eqn is given by,

therefore y_c=c_1cossqrt(2x)+c_2sinsqrt(2x)
For particular integral ,

y_p=1/(f(D))x=1/(D^2+1)e^xcosx+1/(D^2+1)x^2e^(3x)

=e^x1/((D+1)^2+1)cosx+1/(D^2+1)x^2e^(3x)

=e^x1/(D^2+2D+3)cosx+e^(3x)1/((D+3)^2+2)x^2

=e^x1/2(D-1)/(D^2-1)cosx+e^(3x)1/(D^2+6D+11)x^2

=e^x1/4(sinxcosx)+e^(3x)/11[1+(6D+D^2)/11]^(-1)x^2

=e^x1/4(sinxcosx)+e^(3x)/11[1+(6D+D^2)/11+(36D^2)/121+..]x^2

therefore y_p=e^x1/4(sinx+cosx)+e^(3x)/11[x^2-(12x)/11+50/121]

y_g=y_c+y_p=c_1cossqrt(2x)+c_2sinsqrt(2x)+e^x1/4(sinx+cosx)+e^(3x)/11[x^2-(12x)/11+50/121]

Concept: Linear Differential Equation with Constant Coefficientโ Complementary Function
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2017-2018 (December) CBCGS

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