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Question
Solve the following differential equation.
x2y dx − (x3 + y3 ) dy = 0
Solution
x2y dx − (x3 + y3 ) dy = 0
∴ x2y dx = (x3 + y3) dy
∴`dy/dx = (x^2y)/(x^3+y^3)` …(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.tx)/(x^3 + t^3x^3)`
∴ `t + x dt/dx = (x^3.t)/(x^3(1+t^3))`
∴ `x dt/dx = t /(1+ t ^3)-t`
∴ `xdt/dx =( t-t-t^4)/(1+t^3)`
∴ `xdt/dx = (-t^4)/(1+t^3)`
∴ `(1+t^3)/t^4 dt = - dx/x`
Integrating on both sides, we get
`int(1+t^3)/t^4 = - int 1/x dx`
∴ `int(1/t^4+1/t)dt = - int1/x dx`
∴ `int t^-4 dt + int 1/t dt = - int 1/x dx`
∴ ` t^-3/-3+ log |t| = - log |x| + log|c_1|`
∴ ` - 1/-3t^3+ log |t| = - log |x| + log|c_1|`
∴ `-1/3. 1/(y/x)^3+log|y/x| = - log|x| + log|c_1|`
∴ `-x^3/(3y^3)+ log|y|-log|x| = -log|x| + log|c_1|`
∴ `log |y| + log |c| = x^3/(3y^3)` ...[-log |c1| = log |c|]
∴ `log |yc| = x^3/(3y^3)`
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