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Solve the following differential equation. xy dydx=x2+2y2 - Mathematics and Statistics

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प्रश्न

Solve the following differential equation.

`xy  dy/dx = x^2 + 2y^2`

योग
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उत्तर

`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

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 8: Differential Equation and Applications - Exercise 8.4 [पृष्ठ १६७]

APPEARS IN

बालभारती Mathematics and Statistics 1 (Commerce) [English] Standard 12 Maharashtra State Board
अध्याय 8 Differential Equation and Applications
Exercise 8.4 | Q 1.6 | पृष्ठ १६७

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