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प्रश्न
Solve the following differential equation.
`x^2 dy/dx = x^2 +xy - y^2`
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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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