Question

solve (y-x)dy/dx + (2x+3y)=0​

Answer 1

Step-by-step explanation:

We have,

$(y - x) \frac{dy}{dx} + (2x + 3y) = 0$

$\implies(y - x) \frac{dy}{dx} = - (2x + 3y)$

$\implies \frac{dy}{dx} = - \frac{(2x + 3y) }{ (y - x) }$

$\implies \frac{dy}{dx} = \frac{2x + 3y }{ x - y}$

$Let\: \: y=vx$

$\implies\frac{dy}{dx}=v+x\frac{dv}{dx}$

So,

$v + x \frac{dv}{dx} = \frac{2x + 3vx}{ x - vx}$

$\implies \: v + x \frac{dv}{dx} = \frac{x(2 + 3v)}{ x (1- v)}$

$\implies \: v + x \frac{dv}{dx} = \frac{2 + 3v}{1- v}$

$\implies \: x \frac{dv}{dx} = \frac{2 + 3v}{1- v} - v$

$\implies \: x \frac{dv}{dx} = \frac{2 + 3v - v + {v}^{2} }{1- v}$

$\implies \: x \frac{dv}{dx} = \frac{ {v}^{2} + 2v + 2}{1- v}$

$\implies \: \frac{1 - v}{ {v}^{2} + 2v + 2} dv = \frac{1}{x} dx$

$\implies \: \int\frac{1 - v}{ {v}^{2} + 2v + 2} dv = \int\frac{1}{x} dx$

$\implies \: \frac{1}{2} \int\frac{2 - 2v}{ {v}^{2} + 2v + 2} dv = \int\frac{1}{x} dx$

$\implies \: \frac{1}{2} \int\frac{ - 2 - 2v + 2}{ {v}^{2} + 2v + 2} dv = \int\frac{1}{x} dx$

$\implies \: \frac{1}{2} \int\frac{ - 2 - 2v }{ {v}^{2} + 2v + 2} dv + \frac{1}{2} \int \frac{2}{ {v}^{2} + 2v + 2 }dv = \int\frac{1}{x} dx$

$\implies \: - \frac{1}{2} \int\frac{ 2v + 2}{ {v}^{2} + 2v + 2} dv + \int \frac{1}{ {v}^{2} + 2v + 2 }dv = \int\frac{1}{x} dx$

In the first integral, in numerator, derivative of denominator is present,

So,

we have,

$\implies \: - \frac{1}{2} \ln | {v}^{2} + 2v + 2 | + c _{1} + \int \frac{1}{ {v}^{2} + 2v + 1 + 1 }dv = \int\frac{1}{x} dx$

$\implies \: - \frac{1}{2} \ln | {v}^{2} + 2v + 2 | +c _{1} + \int \frac{1}{ {(v + 1)}^{2} + 1 }dv = \ln | x | + c _{3}$

$\implies \: - \frac{1}{2} \ln | {v}^{2} + 2v + 2 | + c _{1}\tan^{ - 1} (v + 1) + c _{2} = \ln | x | + c _{3}$

$\implies \: - \frac{1}{2} \ln | {v}^{2} + 2v + 2 | + \tan^{ - 1} (v + 1) + = \ln | x | + C \: \: \: \: \: \{ \sf{where \: \: C = c_{3} -c_{2} - c_{1}}\}$

$\implies \: - \ln \sqrt{{v}^{2} + 2v + 2 } + \tan^{ - 1} (v + 1) + = \ln | x | + C$

$\implies \: \tan^{ - 1} (v + 1) = \ln | x | + \ln \sqrt{{v}^{2} + 2v + 2 } + C$

Put $v=\frac{y}{x}$

$\implies \: \tan^{ - 1} \bigg( \frac{y}{x} + 1 \bigg) = \ln | x | + \ln \bigg\{ \sqrt{{ \bigg(\frac{y}{x} \bigg) }^{2} + 2 \bigg(\frac{y}{x} \bigg) + 2 } \bigg \} + C$

$\implies \: \tan^{ - 1} \bigg( \frac{x + y}{x} \bigg) = \ln | x | + \ln \bigg\{ \sqrt{ \frac{y ^{2} }{ {x}^{2} } + 2 \frac{y}{x} + 2 } \bigg \} + C$

$\implies \: \tan^{ - 1} \bigg( \frac{x + y}{x} \bigg) = \ln | x | + \ln \bigg\{ \sqrt{ \frac{y ^{2} + 2xy + 2 {x}^{2} }{ {x}^{2} } } \bigg \} + C$

$\implies \: \tan^{ - 1} \bigg( \frac{x + y}{x} \bigg) = \ln | x | + \ln\frac{ \sqrt{y ^{2} + 2xy + 2 {x}^{2} }}{ |x| } + C$

$\implies \: \tan^{ - 1} \bigg( \frac{x + y}{x} \bigg) = \ln | x | + \ln\sqrt{y ^{2} + 2xy + 2 {x}^{2} } - \ln |x| + C$

$\implies \: \tan^{ - 1} \bigg( \frac{x + y}{x} \bigg) = \ln\sqrt{y ^{2} + 2xy + 2 {x}^{2} } + C$