Question

Differential equations of (4y+X)Dy/DX,=3x-y

Answer 1

Step-by-step explanation:

We have,

$(4y + x) \frac{dy}{dx} = 3x - y$

$\implies\frac{dy}{dx} = \frac{3x - y }{4y + x}$

$Let \: y = vx \\ \implies \: \frac{dy}{dx} = v + x \frac{dv}{dx}$

So,

$\implies \: v + x\frac{dv}{dx} = \frac{3x - vx }{4vx + x}$

$\implies \: v + x\frac{dv}{dx} = \frac{x(3 - v)}{x(4v + 1)}$

$\implies \: v + x\frac{dv}{dx} = \frac{(3 - v)}{(4v + 1)}$

$\implies \: x\frac{dv}{dx} = \frac{(3 - v)}{(4v + 1)} - v$

$\implies \: x\frac{dv}{dx} = \frac{(3 - v) - v(4v + 1)}{(4v + 1)}$

$\implies \: x\frac{dv}{dx} = \frac{3 - v - 4v^{2} - v}{4v + 1}$

$\implies \: x\frac{dv}{dx} = \frac{3 - 2v - 4v^{2} }{4v + 1}$

$\implies \frac{(4v + 1)dv}{3 - 2v - 4 {v}^{2} } = \frac{dx}{x}$

$\implies \int\frac{(4v + 1)dv}{3 - 2v - 4 {v}^{2} } = \int\frac{dx}{x}$

$\implies - \frac{1}{2} \int\frac{ - 2(4v + 1)dv}{3 - 2v - 4 {v}^{2} } = \int\frac{dx}{x}$

$\implies \: - \frac{1}{2} ln |3 - 2v - 4 {v}^{2} | = ln |x| + ln(c)$

Here, ln(c) is a integration constant

$\implies \: - \frac{1}{2} ln |3 - 2 \frac{y}{x} - 4 { (\frac{y}{x} )}^{2} | = ln |x| + ln(c)$

$\implies \: - \frac{1}{2} ln |3 - 2 \frac{y}{x} - \frac{4y ^{2} }{x^{2} } | = ln |x| + ln(c)$

$\implies \: - \frac{1}{2} ln |3 {x}^{2} - 2yx -4y ^{2} | + \frac{1}{2} ln |x^{2} | = ln |x| + ln(c)$

$\implies \: - \frac{1}{2} ln |3 {x}^{2} - 2yx -4y ^{2} | + ln |x | = ln |x| + ln(c)$

$\implies \: - ln \sqrt{3 {x}^{2} - 2yx -4y ^{2} } = ln(c)$

$\implies \: ln(c) + ln \sqrt{3 {x}^{2} - 2yx -4y ^{2} } = 0$

$\implies \: ln \{c\sqrt{3 {x}^{2} - 2yx -4y ^{2} } \} = 0$

$\implies \: c\sqrt{3 {x}^{2} - 2yx -4y ^{2} } = 1$

$\implies \: \sqrt{3 {x}^{2} - 2yx -4y ^{2} } = \frac{1}{c}$

$\implies \:3 {x}^{2} - 2yx -4y ^{2} = \frac{1}{ {c}^{2} }$

$\implies \:3 {x}^{2} - 2yx -4y ^{2} = C$

Here C= 1/c²