Question

Which of the following functions is a solution to the differential equation: -6y' + 18x2 = 0 a) Oy= -18c-2 b) y=z? - 6 c) y=x+4 d -6 x² + x e) = (x - 4)

Answer 1

`y = x^3 +C`

Explanation:

Given:

`\text{Solve,} -6y'+18x^2=0.`

`\Longrightarrow -6y'+18x^2=0\\\\\Longrightarrow [-6y'+18x^2=0]-(1)/(6) \\\\\Longrightarrow y'-3x^2=0\\\\\Longrightarrow (dy)/(dx) =3x^2\\`

`\boxed{\left\begin{array}{ccc}\text{\underline{Using Speration of Varibles:}}\\(dy)/(dx)=f(x)g(y) \\\Rightarrow \int(dy)/(g(y))=\int f(x)gx \end{array}\right}`

`\Longrightarrow (dy)/(dx) =3x^2\\\\\Longrightarrow dy =3x^2dx\\\\\Longrightarrow \int dy =\int 3x^2dx\\\\\Longrightarrow \boxed{\boxed{y = x^3 +C}} \therefore Sol.`

Thus, the given first-order differential equation is solved.