Question

Solve the following initial value problem:
(y⁴ − 2x³ y) dx + (x⁴ − 2xy³) dy = 0, y (1) = 1

Answer 1

(y⁴ − 2x³ y) dx + (x⁴ − 2xy³) dy = 0, y (1) = 1
This is an homogenous equation, put y= vx
\[\left( v^4 x^4 - 2v x^4 \right) + \left( x^4 - 2 v^3 x^4 \right) \left[ v + x\frac{dv}{dx} \right] = 0\]
\[\left( v^4 x^4 - 2v x^4 \right) = \left( 2 v^3 x^4 - x^4 \right) \left[ v + x\frac{dv}{dx} \right]\]
\[v x^4 \left( v^3 - 2 \right) = x^4 \left( 2 v^3 - 1 \right) \left[ v + x\frac{dv}{dx} \right]\]
\[v\left( v^3 - 2 \right) = \left( 2 v^3 - 1 \right)v + x\left( 2 v^3 - 1 \right)\frac{dv}{dx}\]
\[v\left[ v^3 - 2 - 2 v^3 + 1 \right] = x\left( 2 v^3 - 1 \right)\frac{dv}{dx}\]
\[v\left( - 1 - v^3 \right) = x\left( 2 v^3 - 1 \right)\frac{dv}{dx}\]
\[v\left( 1 + v^3 \right) = x\left( 1 - 2 v^3 \right)\frac{dv}{dx}\]
\[\frac{dx}{x} = \frac{\left( 1 - 2 v^3 \right)}{v\left( 1 + v^3 \right)}dv\]
On integrating both side of the equation we get,
\[\int\frac{dx}{x} = \int\frac{\left( 1 - 2 v^3 \right)}{v\left( 1 + v^3 \right)}dv\]
\[ \Rightarrow \log_e x = \int\frac{1 + v^3 - 3 v^3}{v\left( 1 + v^3 \right)}dv\]
\[ \Rightarrow \log_e x = \int\frac{1 + v^3}{v\left( 1 + v^3 \right)}dv - \int\frac{3v}{v\left( 1 + v^3 \right)}dv\]
\[ \Rightarrow \log_e x = \int\frac{1}{v}dv - \int\frac{3 v^2}{\left( 1 + v^3 \right)}dv\]
\[ \Rightarrow \log_e x = \log_e v - \int\frac{dt}{t}\]
\[ \Rightarrow \log_e x = \log_e v - \log_e \left( 1 + v^3 \right) + c.......\text{ let }\left( 1 + v^3 \right) = t, 3 v^2 dv = dt\]
\[ \Rightarrow \log_e x = \log_e \frac{v}{1 + v^3} + c\]
As `v = y/x`
\[ \Rightarrow \log_e x = \log_e \frac{\frac{y}{x}}{1 + y^\frac{3}{x}} + c\]
\[ \Rightarrow \log_e x = \log_e \frac{y x^2}{x^3 + y^3} + c\]
As y(1) = 1
\[ \Rightarrow \log_e 1 = \log_e \frac{1}{1 + 1} + c\]
\[ \Rightarrow 0 = \log_e \frac{1}{2} + c\]
\[c = - \log_e \frac{1}{2}\]
\[ \Rightarrow c = \log_e 2\]
\[ \therefore \log_e x = \log_e \frac{y x^2}{x^3 + y^3} + \log_e 2\]