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Solve this differential equation: f''(x) = x^{-3/2}, f'(4) = 1, f(0) = 0

Question:

Solve this differential equation:

{eq}f''(x) = x^{-{3\over 2}},\ f'(4) = 1,\ f(0) = 0 {/eq}

Power Rule in Integration

For an integrand in the form {eq}x^n {/eq}, where {eq}n {/eq} is constant real number, the integral of this is computed by the power rule. The power rule for integrals is defined as

$$\displaystyle \int x^n \,dx = \frac{x^{n+1}}{n+1} $$

Answer and Explanation:

The given differential equation is

{eq}f''(x) = x^{-\frac{3}{2}} {/eq}

To get the first derivative, we will integrate this equation.

{eq}\displaystyle \frac{df'(x)}{dx} = x^{-\frac{3}{2}} \\ \displaystyle df'(x) = x^{-\frac{3}{2}}\,dx \\ \displaystyle \int df'(x) = \int x^{-\frac{3}{2}}\,dx \\ \displaystyle f'(x) = \int x^{-\frac{3}{2}}\,dx {/eq}

Apply the power rule: {eq}\displaystyle \int x^n \,dx = \frac{x^{n+1}}{n+1} {/eq}.

{eq}\displaystyle f'(x) = \frac{x^{-\frac{3}{2}+1}}{-\frac{3}{2}+1} \\ \displaystyle f'(x) = \frac{x^{-\frac{3}{2}+\frac{2}{2}}}{-\frac{3}{2}+\frac{2}{2}} \\ \displaystyle f'(x) = \frac{x^{-\frac{1}{2}}}{-\frac{1}{2}} \\ \displaystyle f'(x) = -\frac{2x^{-\frac{1}{2}}}{1} \\ \displaystyle f'(x) = -2x^{-\frac{1}{2}} {/eq}

Add a constant of integration.

{eq}\displaystyle f'(x) = -2x^{-\frac{1}{2}} + c_1 {/eq}

To get the value of the unknown constant, we will use the given condition, {eq}f'(4) = 1 {/eq}.

{eq}\displaystyle f'(4) = -2(4)^{-\frac{1}{2}} + c_1 = 1 \\ \displaystyle -\frac{2}{(4)^{\frac{1}{2}}} + c_1 = 1 \\ \displaystyle -\frac{2}{\sqrt{4}} + c_1 = 1 \\ \displaystyle -\frac{2}{2} + c_1 = 1 \\ \displaystyle -1 + c_1 = 1 \\ \displaystyle c_1 = 1 + 1 \\ \displaystyle c_1 = 2 {/eq}

So the first derivative is

{eq}\displaystyle f'(x) = -2x^{-\frac{1}{2}} + 2 {/eq}

We will integrate the first derivative to get the solution.

{eq}\displaystyle \frac{df}{dx} = -2x^{-\frac{1}{2}} + 2 \\ \displaystyle df = (-2x^{-\frac{1}{2}} + 2)\,dx \\ \displaystyle \int df = \int (-2x^{-\frac{1}{2}} + 2)\,dx \\ \displaystyle f(x) = \int (-2x^{-\frac{1}{2}} + 2)\,dx {/eq}

Apply the sum rule: {eq}\displaystyle \int (f(x)\pm g(x))\,dx = \int f(x)\,dx \pm \int g(x)\,dx {/eq}.

{eq}\displaystyle f(x) = \int -2x^{-\frac{1}{2}}\,dx + \int 2\,dx {/eq}

Take out the constants.

{eq}\displaystyle f(x) = -2\int x^{-\frac{1}{2}}\,dx + 2\int dx \\ \displaystyle f(x) = -2\int x^{-\frac{1}{2}}\,dx + 2x {/eq}

Apply power rule.

{eq}\displaystyle f(x) = -2\left ( \frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1} \right ) + 2x \\ \displaystyle f(x) = -2\left ( \frac{x^{-\frac{1}{2}+\frac{2}{2}}}{-\frac{1}{2}+\frac{2}{2}} \right ) + 2x \\ \displaystyle f(x) = -2\left ( \frac{x^{\frac{1}{2}}}{\frac{1}{2}} \right ) + 2x \\ \displaystyle f(x) = -2\left ( \frac{2x^{\frac{1}{2}}}{1} \right ) + 2x \\ \displaystyle f(x) = -4x^{\frac{1}{2}} + 2x {/eq}

Add a constant of integration.

{eq}\displaystyle f(x) = -4x^{\frac{1}{2}} + 2x + c_2 {/eq}

To get the value of the constant, we will use the condition {eq}f(0) = 0 {/eq}.

{eq}\displaystyle f(0) = -4(0)^{\frac{1}{2}} + 2(0) + c_2 = 0 \\ \displaystyle -4\sqrt{0} + 0 + c_2 = 0 \\ \displaystyle -4(0) + c_2 = 0 \\ \displaystyle 0 + c_2 = 0 \\ \displaystyle c_2 = 0 {/eq}

Hence, the solution of the differential equation is

{eq}\displaystyle \boldsymbol{f(x) = 2x - 4x^{\frac{1}{2}}} {/eq}


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Integration Problems in Calculus: Solutions & Examples

from AP Calculus AB & BC: Homework Help Resource

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