Solve the initial value problem
y"-3y'-10y=0, y'(0)=1, y'(0)=2
y"−3y′−10y=0,y(0)=1,y′(0)=2y"-3y'-10y=0, y(0)=1, y'(0)=2y"−3y′−10y=0,y(0)=1,y′(0)=2
Auxiliary equation is
k2−3k−10=0k^2-3k-10=0k2−3k−10=0
k=3±9+402k=\frac{3\pm \sqrt{9+40}}{2}k=23±9+40
k1=−2,k2=5k_1=-2,k_2=5k1=−2,k2=5
General solution is
y=c1e−2x+c2e5x...(1)y=c_1e^{-2x}+c_2e^{5x}...(1)y=c1e−2x+c2e5x...(1)
Since, y(0)=1y(0)=1y(0)=1
1=c1+c2...(2)1=c_1+c_2...(2)1=c1+c2...(2)
And
y′=−2c1e−2x+5c2e5xy'=-2c_1e^{-2x}+5c_2e^{5x}y′=−2c1e−2x+5c2e5x
Also y′(0)=2y'(0)=2y′(0)=2
2=−2c1+5c2...(3)2=-2c_1+5c_2...(3)2=−2c1+5c2...(3)
Multiply equation (2) with 2 and add with equation (3), we get
4=7c2⇒c2=474=7c_2\Rightarrow c_2=\frac{4}{7}4=7c2⇒c2=74
From equation (2),
1=c1+47⇒c1=371=c_1+\frac 4 7\Rightarrow c_1=\frac 3 71=c1+74⇒c1=73
y=37e−2x+47e5x⇒7y=3e−2x+4e5xy=\frac 3 7e^{-2x}+\frac 4 7e^{5x}\Rightarrow 7y=3e^{-2x}+4e^{5x}y=73e−2x+74e5x⇒7y=3e−2x+4e5x
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments