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**Test 3 / MAT334 TT3 MAIN A Q2**

« **on:**December 07, 2020, 03:46:01 PM »

Hello everyone, here is the answer of MAT334 TT3 MAIN A Q2

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Hello everyone, here is the answer of MAT334 TT3 MAIN A Q2

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Hello everyone , here is the Q1 of TT3 Q1 Main A

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hi everyone here is the quiz question find the general solution of the given differential equation

2y'' -3y' +y = 0

therefore 2r^2 -3r + 1 = 0

therefore (2r-1)(r-1) = 0

R1 = 1/2 R2 = 1

therefore y = C1e^((1/2)t)+C2e^t

2y'' -3y' +y = 0

therefore 2r^2 -3r + 1 = 0

therefore (2r-1)(r-1) = 0

R1 = 1/2 R2 = 1

therefore y = C1e^((1/2)t)+C2e^t

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here is a quesion form quiz 2 everyone enjoy!

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Find the general solution of the given equation:

xy' = (1-y^2)^1/2

solution: Because Separable,

therefore x(dy/dx) = (1-y^2)^1/2,

Rearrange: \int(1/(1-y^2)^1/2)*dy =\int(1/x)dx, where x not equal 0, y does not equal to positive or negative 1,

therefore: arcsin(y) = ln|x| + C,

therefore general solution: y=sin(ln|x| + C) where x not equal to 0, y not equal to positive 1 or negative 1.

xy' = (1-y^2)^1/2

solution: Because Separable,

therefore x(dy/dx) = (1-y^2)^1/2,

Rearrange: \int(1/(1-y^2)^1/2)*dy =\int(1/x)dx, where x not equal 0, y does not equal to positive or negative 1,

therefore: arcsin(y) = ln|x| + C,

therefore general solution: y=sin(ln|x| + C) where x not equal to 0, y not equal to positive 1 or negative 1.

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