MTH401 ASSIGNMENT NO. 1 FALL 2022 || 100% RIGHT SOLUTION || DIFFERENTIAL EQUATIONS || BY VuTech

MTH401 ASSIGNMENT NO. 1 FALL 2022

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MTH401 ASSIGNMENT NO. 1 FALL 2022 || 100% RIGHT SOLUTION || DIFFERENTIAL EQUATIONS || BY VuTech

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Assignment # 1                                      MTH401 (Fall 2022)

 Total Marks: 10                                                                               Due Date: December 5, 2022.

 

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QUESTION 1: 

Solve the Differential equation

`(3x^2 + 9xy + 5y^2)dx - (6x^2 + 4xy)dy = 0); y(2) = -6` 

SOLUTION:

`(3x^2 + 9xy + 5y^2)dx - (6x^2 + 4xy)dy = 0`

`(3x^2 + 9xy + 5y^2)dx = (6x^2 + 4xy)dy`

`\frac{(3x^2 + 9xy + 5y^2)}{(6x^2 + 4xy)} = \frac{dy}{dx}`

`\frac{x^2(3 + 9(\frac{y}{x}) + 5(\frac{y}{x})^2)}{x^2(6 + 4(\frac{y}{x}))} = \frac{dy}{dx}`

`\frac{3 + 9(\frac{y}{x}) + 5(\frac{y}{x})^2}{6 + 4(\frac{y}{x})} = \frac{dy}{dx}`

`\frac{dy}{dx} = \frac{3 + 9(\frac{y}{x}) + 5(\frac{y}{x})^2}{6 + 4(\frac{y}{x})}            -  -  -  -  -  -  - (1)`

Now

`y = vx`

`v = \frac{y}{x}`

Put the value of  `\frac{dy}{dx},` in (1)

`v + x\frac{dv}{dx} = \frac{3 + 9v + 5v^2}{6 + 4v}`

`x\frac{dv}{dx} = \frac{3 + 9v + 5v^2}{6 + 4v} - v`

`x\frac{dv}{dx} = \frac{3 + 9v + 5v^2 - v(6 + 4v)}{6 + 4v}`

`x\frac{dv}{dx} = \frac{3 + 9v + 5v^2 - 6v - 4v^2}{6 + 4v}`

`x\frac{dv}{dx} = \frac{v^2 + 3v + 3}{6 + 4v}`

`\frac{1}{x}\frac{dx}{dv} = \frac{6 + 4v}{v^2 + 3v + 3}`

`\frac{dx}{x} = (\frac{6 + 4v}{v^2 + 3v + 3})dv`

Apply Integral on both sides

`\int \frac{dx}{x} = \int (\frac{6 + 4v}{v^2 + 3v + 3})dv `

`\int \frac{dx}{x} = \int (\frac{2(3 + 2v)}{v^2 + 3v + 3})dv`

`\int \frac{dx}{x} = \int 2(\frac{3 + 2v}{v^2 + 3v + 3})dv  `

`\ln |x| + \ln |c| = 2\ln |(v^2 + 3v + 3)|`

`\ln |xc| = \ln |(v^2 + 3v + 3)^2|`

`xc = (v^2 + 3v + 3)^2`

By putting `v = \frac{y}{x}`

`xc = (\frac{y}{x})^2} + 3\frac{y}{x} + 3)^2`

`xc = (\frac{y}{x^2}^2 + 3\frac{y}{x} + 3)^2`

By taking L.C.M

`xc = \frac{(y^2 + 3xy + 3x^2)^2}{(x^2)^2}`

`xc = \frac{(y^2 + 3xy + 3x^2)^2}{x^4}`

`x^4.xc = (y^2 + 3xy + 3x^2)^2`

`x^{4 + 1}c = (y^2 + 3xy + 3x^2)^2`

`x^5c = (y^2 + 3xy + 3x^2)^2`

By Putting y(2) = - 6

 

`2^5c = (( - 6)^2 + 3(2)( - 6) + 3(2^2))^2`

`32c = (36 - 36 + 12)^2`

`32c = (12)^2`

`32c = 144`

`c = \frac{144}{32}`

`c = \frac{9}{2}`

Now put `c = \frac{9}{2}` in this equation `x^5c = (y^2 + 3xy + 3x^2)^2`

`x^5\frac{9}{2} = (y^2 + 3xy + 3x^2)^2`

`\frac{9x^5}{2} = (y^2 + 3xy + 3x^2)^2`

  

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