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### Topics - Dang Tongbo

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##### Quiz-3 / Tut 0301 - Quiz 3
« on: October 13, 2019, 03:07:39 PM »
2y'' + 4y' -4y = 0 , y(0) = 0, y'(0) = 1.
Solution:
2(r^2)+r-4= 0
so, r1 = ((-1)+(33)^(1/2))/4
r2 = ((-1)+(33)^(1/2))/4
y = c1*e^(((-1)+(33)^(1/2))/4)+c2*e^(((-1)-(33)^(1/2))/4).
Because y(0)=0, so, c1+c2 = 0.
y' = -((-1)-(33)^(1/2))/4)*c1*e^(((-1)+(33)^(1/2))/4)-((1+(33)^1/2))*c2*e^(((-1)-(33)^(1/2))/4)
Because y'(0) = 1, so 1 = -(1-(33)^(1/2)/4)*c1 - (1+(33)^(1/2))/4)*c2
c1 = -c2,
so, -(1-(33)^(1/2)/4)*(-c2) - (1+(33)^(1/2))/4)*c2=1
c2 = -(4/(2*(33)^1/2))
c1 = 4/(2*(33)^1/2)
so, y = 4/(2*(33)^1/2)*e^((-1-(33)^1/2))/4)*t - (4/2*(33)^(1/2))/4)*t

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##### Quiz-2 / Tut 5103- Quiz 2
« on: October 06, 2019, 05:23:05 PM »
Question: (x+2)* sin(y) + x*cox(y)*(dy/dx) = 0, u(x,y) = x*e^(x)
Solution:M(x,y) = (x+2)*sin(y),N(x,y) = cos(y)
My = (x+2)*cos(y), Nx= cos(y)
Because My is not equal to Nx, the equation is not exact.
In this way, (x+2)*(x*e^(x))*sin(y)+ x^2 *(e^x)*cos(y)*(dy/dx)=0
so, Lx(x,y) = (x+2)x(e^x)sin(y)
Ly(x,y) = (x^2)(e^x)cos(y)
L(x,y) = int((x+2)x*(e^x)sin(y)dx = (x^2)^(e^x)sin(y)+h(y)
Ly(x,y)= (x^2)(e^x)cos(y)+h'(y)= (x^2)(e^x)cos(y)
h'(y) = 0
As a result, (x^2)*(e^x)sin(y)=C

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