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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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