Is $y = \frac { 1 } { 12 } x ^ { 4 } - \frac { 5 } { 2 } x ^ { 3 } + x ^ { 2 } + C _ { 1 } x + C _ { 2 }$ the general solution to the differential equation $\frac { d ^ { 2 } y } { d x ^ { 2 } } = x ^ { 2 } - 15 x + 2$ ?

Question

Quizplus · Accepted Answer

To verify if $y = \frac { 1 } { 12 } x ^ { 4 } - \frac { 5 } { 2 } x ^ { 3 } + x ^ { 2 } + C _ { 1 } x + C _ { 2 }$ is the general solution to the given differential equation, we need to differentiate $y$ twice with respect to $x$ and see if we obtain the right-hand side of the differential equation, $\frac { d ^ { 2 } y } { d x ^ { 2 } } = x ^ { 2 } - 15 x + 2$.First derivative of $y$ with respect to $x$:$$ \frac{dy}{dx} = \frac{d}{dx} \left( \frac { 1 } { 12 } x ^ { 4 } - \frac { 5 } { 2 } x ^ { 3 } + x ^ { 2 } + C _ { 1 } x + C _ { 2 } \right) = \frac{1}{3}x^3 - \frac{15}{2}x^2 + 2x + C_1 $$Second derivative of $y$ with respect to $x$:$$ \frac{d^2y}{dx^2} = \frac{d}{dx} \left( \frac{1}{3}x^3 - \frac{15}{2}x^2 + 2x + C_1 \right) = x^2 - 15x + 2 $$This matches the given differential equation, confirming that the provided $y$ is indeed its general solution.

Is y=112x4−52x3+x2+C1x+C2y = \frac { 1 } { 12 } x ^ { 4 } - \frac { 5 } { 2 } x ^ { 3 } + x ^ { 2 } + C _ { 1 } x + C _ { 2 }y=121​x4−25​x3+x2+C1​x+C2​

Is $y = \frac { 1 } { 12 } x ^ { 4 } - \frac { 5 } { 2 } x ^ { 3 } + x ^ { 2 } + C _ { 1 } x + C _ { 2 }$