Calculus
Please look at my work below: Solve the initial-value problem. y'' + 4y' + 6y = 0 , y(0) = 2 , y'(0) = 4 r^2+4r+6=0, r=(16 +/- Sqrt(4^2-4(1)(6)))/2(1) r=(16 +/- Sqrt(-8)) r=8 +/- Sqrt(2)*i, alpha=8, Beta=Sqrt(2) y(0)=2, e^(8*0)*(c1*cos(0)+c2*sin(0))=c2=2 y'...
Calculus - Second Order Differential Equations
Solve the initial-value problem. y'' - 2y' + y = 0 , y(2) = 0 , y'(2) = 1 r^2-2r+1=0, r1=r2=1 y(x)=c1*e^x+c2*x*e^x y(2)=c1*e^2+c2*2*e^2=0 c1=-(2*c2*exp(2))/exp(2) c1=-2*c2 y'(x)=-2*c2*e^x+c2*e^x*(x-1) y'(2)=-2*c2*e^2+c2*e^2*(2-1)=1 c2(-2e^2+e^2)=1 c2=1/...
Calculus - Second Order Differential Equations
Posted by COFFEE on Monday, July 9, 2007 at 9:10pm. download mp3 free instrumental remix Solve the initial-value problem. y'' + 4y' + 6y = 0 , y(0) = 2 , y'(0) = 4 r^2+4r+6=0, r=(16 +/- Sqrt(4^2-4(1)(6)))/2(1) r=(16 +/- Sqrt(-8)) r=8 +/- Sqrt(2)*i, alpha=8, Bet...
Calculus - Second Order Differential Equations
Solve the initial-value problem. y'' + 4y' + 6y = 0 , y(0) = 2 , y'(0) = 4 r^2+4r+6=0, r=(16 +/- Sqrt(4^2-4(1)(6)))/2(1) r=(16 +/- Sqrt(-8)) r=8 +/- Sqrt(2)*i, alpha=8, Beta=Sqrt(2) y(0)=2, e^(8*0)*(c1*cos(0)+c2*sin(0))=c2=2 y'(0)=4, c2=4 y(x)=e^(8x)*(2*cos...
Calculus - Second Order Differential Equations
Solve the boundary-value problem. y''+5y'-6y=0, y(0)=0, y(2)=1 r^2+5r-6=0, r1=1, r2=-6 y=c1*e^x + c2*e^-6x y(x)=c1*e^x+c2*e^-6x y'(x)=c1*e^x-6*c2*e^-6x y(0)=c1+c2=0, c1=-c2 y(2)=c1*e^2+c2*e^(-12)=1 -c2*e^2-6c2*e^(-12)=1 -c2(e^2-6*e^-12)=1 c2=-1/(e^2-6e^-12) c1=...
calc check please?
Given the differential equation: dy/dx = y(1+x), y(0)=1, Use Euler's method with step size .1 to approximate y(.3). ... please check this for me! no one has responded to this question yet.. thanks. y' = y(1+x), y'(0) = 1(1+0)=1 ->the solution has slope 1 at the ...
calc: avg value
Find the average value of the function "f(x) = x^2 sqrt(1+x^3)" on the interval [0,2]. and this is what i did.. please check for mistakes. thanks :D f(x) = x^2 sqrt(1+x^3), [0,2] f ave = (1/(b-a))*inegral of a to b for: f(x) dx f ave = (1/(2-0))*integral of 0 to 2 f...
calc check: curve length
Find the length of the curve y=(1/(x^2)) from ( 1, 1 ) to ( 2, 1/4 ) [set up the problem only, don't integrate/evaluate] this is what i did.. let me know asap if i did it right.. y = (1/(x^2)) dy/dx = (-2/(x^3)) L = integral from a to b for: sqrt(1+(dy/dx)^2)dx L = integra...
calc check: euler's method
Given the differential equation: dy/dx = y(1+x), y(0)=1, Use Euler's method with step size .1 to approximate y(.3). y' = y(1+x), y'(0) = 1(1+0)=1 ->the solution has slope 1 at the point (0,1); x0=0, y0=1, h=0.1, F(x,y)=y(1+x) y1=y0+h*F(x0,y0) y1=1 + 0.1(1(1+0)) ...
calc check: average value
Find the average value of the function "f(x) = x^2 sqrt(1+x^3)" on the interval [0,2]. and this is what i did.. please check for mistakes. thanks :D f(x) = x^2 sqrt(1+x^3), [0,2] f ave = (1/(b-a))*inegral of a to b for: f(x) dx f ave = (1/(2-0))*integral of 0 to 2 fo...
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