y"−3y′−10y=0,y(0)=1,y′(0)=2
Auxiliary equation is
k2−3k−10=0
k=23±9+40
k1=−2,k2=5
General solution is
y=c1e−2x+c2e5x...(1)
Since, y(0)=1
1=c1+c2...(2)
And
y′=−2c1e−2x+5c2e5x
Also y′(0)=2
2=−2c1+5c2...(3)
Multiply equation (2) with 2 and add with equation (3), we get
4=7c2⇒c2=74
From equation (2),
1=c1+74⇒c1=73
y=73e−2x+74e5x⇒7y=3e−2x+4e5x
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