Question

Solve the differential equation d2y/dx2=a+bx+cx2 given that dy/dx=0 and y=d when x=0.

Answer 1

Answer to the Question #87111 – Math – Differential Equations

Question

Solve the differential equation

$$d2y/dx^2 = a + bx + cx^2$$

given that

$$dy/dx = 0$$

and

$$y = d$$

when

$$x = 0$$

.

Solution

We want to solve the following differential equation:

$$\frac{d^2 y}{dx^2} = a + bx + cx^2, \quad \left. \frac{dy}{dx} \right|_{x=0} = 0, \quad y(x=0) = d.$$

By integrating both sides of the equation and applying the Second Fundamental Theorem of Calculus we have

$$\int_0^t \frac{d^2 y}{dx} dx = \int_0^t a + bx + cx^2 dx \Rightarrow$$ $$\begin{aligned} \frac{dy}{dx} \big|_{x=t} - \frac{dy}{dx} \big|_{x=0} &= \left( ax + \frac{1}{2} bx^2 + \frac{1}{3} cx^3 \right) \big|_{x=t} - \left( ax + \frac{1}{2} bx^2 + \frac{1}{3} cx^3 \right) \big|_{x=0} \\ &\Rightarrow \frac{dy}{dt} = at + \frac{1}{2} bt^2 + \frac{1}{3} ct^3. \end{aligned}$$

By doing the same process on the last equation we get

$$\int_0^s \frac{dy}{dt} dt = \int_0^s at + \frac{1}{2} bt^2 + \frac{1}{3} ct^3 dt \Rightarrow$$ $$\begin{aligned} y(s) - y(0) &= \left( \frac{1}{2} at^2 + \frac{1}{6} bt^3 + \frac{1}{12} ct^4 \right) \big|_{t=s} - \left( \frac{1}{2} at^2 + \frac{1}{6} bt^3 + \frac{1}{12} ct^4 \right) \big|_{t=0} \\ &\Rightarrow y(s) = \frac{1}{2} as^2 + \frac{1}{6} bs^3 + \frac{1}{12} cs^4 + d. \end{aligned}$$

Please note that in the above calculations we have used the initial conditions given in the question. Thus, the solution of the differential equation is

$$y = \frac{1}{2} ax^2 + \frac{1}{6} bx^3 + \frac{1}{12} cx^4 + d.$$

Answer provided by https://www.AssignmentExpert.com