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Solve the following differential equation. x2dydx=x2+xy-y2 - Mathematics and Statistics

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प्रश्न

Solve the following differential equation.

`x^2 dy/dx = x^2 +xy - y^2`

बेरीज
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उत्तर

`x^2 dy/dx = x^2 +xy - y^2`

∴ `dy/dx = (x^2 + xy - y^2)/x^2`  …(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 + x(tx) - (tx)^2)/x^2`

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

∴ `t+x dt/dx = 1 +t -t^2`

∴  `x dt/dx = 1 + t - t^2`

∴  `x dt/dx = 1 -t^2`

∴  `dt/(1^2- t^2) = dx/x`

Integrating on both sides, we get

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

∴ `1/(2(1))  log  | (1+t)/(1-t)| = log  |x| + log |c_1|`  ...[`Qint dx/(a^2 - x^2) = 1/(2a)  log | (a+x)/(a-x)| +c]`

∴ `log |(1+t)/(1-t)|= log |x| + log|c_1|`

∴ `log |(1+t)/(1-t)|= log |c_1^2x^2|`

∴ `(1+(y/x))/(1-(y/x)) = c_1^2x^2`

∴ `(x+y)/(x-y) = cx^2 … [c_1^2 = c]`

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पाठ 8: Differential Equation and Applications - Exercise 8.4 [पृष्ठ १६७]

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बालभारती Mathematics and Statistics 1 (Commerce) [English] Standard 12 Maharashtra State Board
पाठ 8 Differential Equation and Applications
Exercise 8.4 | Q 1.7 | पृष्ठ १६७

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