How to solve this equation? $a^2 \cdot \arcsin(\frac 4a)+4 \cdot\sqrt {a^2-16}=40$

Question

Please, help me solve this equation: $$a^2 \cdot \arcsin \left(\frac 4a \right)+4 \cdot\sqrt {a^2-16}=40$$

Answer 1

Say $\frac{4}{a}=u$

Then the given equation becomes $$\arcsin u + u\sqrt{1-u^2}=\frac{5}{2}u^2$$

Now $u$ must be such that $|u|<1$.

So we can assume that $u=\sin x$ or $\cos x$.

Now see if you can solve this trigonometric equation.

Answer 2

There may not exist an explicit solution for $a$ in terms of elementary functions.

However, you can apply the Newton-Raphson Method (A numerical method) using a spreadsheet, or with more sophisticated software such as MATLAB:

$$a_{n+1}=a_n-\frac{f(a_n)}{f'(a_n)} \tag{1}$$

You have your function:

$$f(a)=a^2 \cdot \arcsin \left(\frac 4a \right)+4 \cdot\sqrt {a^2-16}-40$$

Differentiating it:

$$f'(a)=\frac{4a}{\sqrt{a^2-16}}-\frac{4}{\sqrt{1-\frac{16}{a^2}}}+2a\arcsin\left(\frac{4}{a}\right)$$

Substituting into $(1)$:

$$a_{n+1}=a_n-\frac{a_n^2 \cdot \arcsin \left(\frac {4}{a_n} \right)+4 \cdot\sqrt {a_n^2-16}-40}{\frac{4a_n}{\sqrt{a_n^2-16}}-\frac{4}{\sqrt{1-\frac{16}{a_n^2}}}+2a_n\arcsin\left(\frac{4}{a_n}\right)} \tag{2}$$

Let us start with an initial value $a_0=11$. You can choose these values by simply 'guessing' a value close to the actual solution.

Now, substitute the value of $a_0$ for $a_n$, and find the value of $a_1$ using equation $(2)$. Continue this iteration until you get closer and closer to the value of $a$.

I get a value of $a\approx 5.52978022$.

Answer 3

If you don't care much about precision, then you can get an approximate solution by fixed point iteration instead of Newton-Raphson.

Using @SchrodingersCat's result, we have an equation:

$$u=\frac{2}{5} \left(\frac{\arcsin u}{u} +\sqrt{1-u^2}\right)$$

Take initial guess $u_0=1$ and substitute:

$$u_1=\frac{2}{5} \left(\frac{\arcsin u_0}{u_0} +\sqrt{1-u_0^2}\right)$$

$$u_2=\frac{2}{5} \left(\frac{\arcsin u_1}{u_1} +\sqrt{1-u_1^2}\right)$$

Then we get the following sequence of values (correct digits are highlighted):

$$\color{blue}{0}.6283185307179586$$ $$\color{blue}{0.7}436955340482698$$ $$\color{blue}{0.7}184422024307543$$ $$\color{blue}{0.72}45114408003248$$ $$\color{blue}{0.723}0826224901516$$ $$\color{blue}{0.723}4206633738902$$ $$\color{blue}{0.7233}407802435841$$ $$\color{blue}{0.72335}96627961654$$ $$\color{blue}{0.72335}51996817168$$ $$\color{blue}{0.723356}2546078128$$ $$\color{blue}{0.7233560}052606297$$ $$\color{blue}{0.7233560}641975273$$ $$\color{blue}{0.72335605}02669220$$ $$\color{blue}{0.72335605}35596262$$

Thus, $u \approx 0.72335605$, so $$a=\frac{4}{u} \approx 5.52978$$