Answer:
[tex]4y+ y"x^2-4y'x = 0[/tex]
↓
[tex]F(x, y, y', y'') = 0[/tex]
Step-by-step explanation:
From the information given:
[tex]y = c_1x + c_2x^4 \\ \\ y^1'= c_1 + 4c_2x^3 \\ \\ c_1 =y' - 4c_2x^3 \\ \\[/tex]
So;
[tex]y = y'x -4c_2x^4+c_2x^4 \\ \\ y - y'x - 3c_2x^4 \\ \\ y'= y"x+y'- 12c_2x^3 \\ \\ c_2= \dfrac{y"x}{12x^3} \\ \\ c_2= \dfrac{y"}{12x^2}[/tex]
∴
[tex]y = y'x- \dfrac{3x^4*y'' }{12x^2} \\ \\ \\ y = y'x - \dfrac{y" x^2}{4} \\ \\ \\ 4y+ y"x^2-4y'x = 0[/tex]
↓
[tex]F(x, y, y', y'') = 0[/tex]
