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Quiz-4 / TUT0602 Quiz 4
« on: October 18, 2019, 01:19:23 PM »
Q: 4y''+9y=0
A:
4r^2+9=0
r^2=-9/4
r=3/2*i
λ=0, μ=3/2
y(t) = c1cos(3/2*t)+c2sin(3/2*t)
A:
4r^2+9=0
r^2=-9/4
r=3/2*i
λ=0, μ=3/2
y(t) = c1cos(3/2*t)+c2sin(3/2*t)
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Quiz-3 / TUT0602 Quiz 3
« on: October 11, 2019, 01:19:54 PM »
Q: Find the Wronskian of two solutions of the given differential equation without solving the equation
cos(t)y''+sin(t)y'-ty=0
Answer:
y''+sin(t)y'/cos(t)-ty/cos(t)=0
P(t)=sin(t)/cos(t)
W=ce^(∫p(t)dt)
=ce^(-∫(sin(t)/cos(t))dt) let u=cos(t), du=-sin(t)dt
=ce^(∫1/u du)
=ce^(lnu)
=cu
=ccos(t)
cos(t)y''+sin(t)y'-ty=0
Answer:
y''+sin(t)y'/cos(t)-ty/cos(t)=0
P(t)=sin(t)/cos(t)
W=ce^(∫p(t)dt)
=ce^(-∫(sin(t)/cos(t))dt) let u=cos(t), du=-sin(t)dt
=ce^(∫1/u du)
=ce^(lnu)
=cu
=ccos(t)
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Quiz-2 / TUT0602 Quiz2
« on: October 04, 2019, 03:47:34 PM »
Q: x²y³+x(1+y²)y′=0, µ(x,y)=1/(xy³)
A:
Let M=x²y³, N=x(1+y²)
My=2x²y², Nx=1+y²
Since My≠Nx, this equation is not exact.
Then, multiple both sides by µ(x,y)=1/(xy³)
We have, x+(1/y³+1/y)y′=0
Let A=x, B=1/y³+1/y
Ay=Bx=0, thus, this equation now is exact.
therefore, ∃Ф(x,y), s.t. Фx=A, Фy=B
Фx=A=x
Ф(x,y)=∫xdx=1/2x²+h(y)
Фy=h′(y)=B=1/y³+1/y
h(y)=∫(1/y³+1/y)dy=-1/(2y²)+ln|y|+C
Ф(x,y)=1/2x²-1/(2y²)+ln|y|+C
General solution: 1/2x²-1/(2y²)+ln|y|=C
A:
Let M=x²y³, N=x(1+y²)
My=2x²y², Nx=1+y²
Since My≠Nx, this equation is not exact.
Then, multiple both sides by µ(x,y)=1/(xy³)
We have, x+(1/y³+1/y)y′=0
Let A=x, B=1/y³+1/y
Ay=Bx=0, thus, this equation now is exact.
therefore, ∃Ф(x,y), s.t. Фx=A, Фy=B
Фx=A=x
Ф(x,y)=∫xdx=1/2x²+h(y)
Фy=h′(y)=B=1/y³+1/y
h(y)=∫(1/y³+1/y)dy=-1/(2y²)+ln|y|+C
Ф(x,y)=1/2x²-1/(2y²)+ln|y|+C
General solution: 1/2x²-1/(2y²)+ln|y|=C
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