Question

Find a solution to the initial value problem, y′′+18x=0,y(0)=5,y′(0)=1.

Answer 1

We want to find a solution to the initial value problem:

$y'' + 18x = 0 \qquad,\qquad y(0) = 5 \qquad,\qquad y'(0)=1.$

We can start by integrating the equation once:

$\dfrac{\textrm{d}^2 y}{\textrm{d}x^2} + 18 x = 0 \iff \dfrac{\textrm{d}^2 y}{\textrm{d}x^2} = -18 x \iff\\\\\iff \dfrac{\textrm{d}y}{\textrm{d}x} = -18\displaystyle\int x\textrm{ d}x \iff \dfrac{\textrm{d}y}{\textrm{d}x}=-18\dfrac{x^2}{2} + C \iff\\\\\iff \dfrac{\textrm{d}y}{\textrm{d}x} = -9x^2 + C.$

Using the initial condition $y'(0) = 1$, we can determine the integration constant $C$:

$\dfrac{\textrm{d}y}{\textrm{d}x}\Big\vert_{x= 0} = 1 \iff -9 \times 0^2 + C = 1 \iff C = 1.$

Therefore, we have:

$\dfrac{\textrm{d}y}{\textrm{d}x} = -9x^2 + 1$

We can now integrate again:

$y(x) = \displaystyle\int\dfrac{\textrm{d}y}{\textrm{d}x}\textrm{ d}x = \int\left(-9x^2+1\right)\textrm{d}x = -9\int x^2\textrm{ d}x + \int\textrm{d}x =\\\\= -9\dfrac{x^3}{3} + x + K = -3x^3 + x + K.$

The integration constant $K$ is determined by using $y(0) = 5$:

$y(0) = 5 \iff -3 \times 0^3 + 0 + K = 5 \iff K = 5.$

Finally, the solution is:

$\boxed{y(x) = -3x^3 + x + 5}.$

Answer 2

By separation of variables, the solution is given by:

$y(x) = -3x^3 + x + 5$

The differential equation is:

$y^{\prime\prime}(x) + 18x = 0$

$y^{\prime\prime}(x) = -18x$

Applying separation of variables:

$\int y^{\prime\prime}(x) = -\int 18x dx$

$y^{\prime}(x) = -9x^2 + K$

Since $y^{\prime}(0) = 1, K = 1$

Thus:

$y^{\prime}(x) = -9x^2 + 1$

To find y, another separation of variables is appled:

$\int y^{\prime}(x) = \int(-9x^2 + 1)dx$

$y(x) = -3x^3 + x + K$

Since y(0) = 5, K = 5, thus, the solution is:

$y(x) = -3x^3 + x + 5$

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