Maharashtra State BoardHSC Commerce 12th Board Exam
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Solve the following differential equation. xy dydx=x2+2y2 - Mathematics and Statistics

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Sum

Solve the following differential equation.

`xy  dy/dx = x^2 + 2y^2`

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Solution

`xy  dy/dx = x^2 + 2y^2`

∴ `dy/dx = (x^2 + 2y^2)/(xy) …(i)`

Put y = tx  ...(ii)

Differentiating w.r.t. x, we get

`dy/dx = t + x dt/dx`  ...(iii)

Substituting (ii) and (iii) in (i), we get

`t +x dt/dx = (x^2 + 2t^2 x^2)/(x(tx))`

∴`t +x dt/dx = (x^2 (1 + 2t^2))/(x^2t)`

∴ `x dt/dx = (1 + 2t^2)/t  - t = (1+t^2)/t`

∴ `t / (1+t^2) dt = 1/xdx`

Integrating on both sides, we get

`1/2 int (2t)/(1+t^2) dt = int dx/x`

∴ `1/2 log |1+ t^2| = log|x| + log |c_1|`

∴ log |1 + t2 | = 2 log |x| + 2log |c1|

= log |x2| + log |c|    …[logc12 = log c]

∴ log |1 + t2| = log |cx 2|

∴ 1 + t2 = cx2

∴ `1+ y^2/x^2 = cx^2`

∴ x2 + y2 = cx4

Concept: Differential Equations
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APPEARS IN

Balbharati Mathematics and Statistics 1 (Commerce) 12th Standard HSC Maharashtra State Board
Chapter 8 Differential Equation and Applications
Exercise 8.4 | Q 1.6 | Page 167
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