solve (D^2+4)y=x^2
pls give me the answer
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Answer:
Auxiliary equation is m2–4=(m+2)(m−2)=0
The roots are 2 and -2.
The complimentary function is yc=C1e2x+C2e−2x
Particular integral
yp=1D2–4x2
=1−4(1−D24)x2
=(1−D24)−1−4x2
=(1+D24+D416+….)−1−4x2
=(x2+24+0+….)−4x2
=−(2x2+1)8
y=yc+yp
=C1e2x+C2e−2x−(2x2+1)8
Step-by-step explanation:
Answered by
1
Answer:
Auxiliary equation is m2–4=(m+2)(m−2)=0
The roots are 2 and -2.
The complimentary function is yc=C1e2x+C2e−2x
Particular integral
yp=1D2–4x2
=1−4(1−D24)x2
=(1−D24)−1−4x2
=(1+D24+D416+….)−1−4x2
=(x2+24+0+….)−4x2
=−(2x2+1)8
y=yc+yp
=C1e2x+C2e−2x−(2x2+1)8
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