Question

$4x{y}^{\prime \prime }+2y^{\prime }+y=0$

Short answer

The general solution is $y(x)=C_1{e}^{r_{1}x}+C_2{e}^{r_{2}x}$.

Step-by-step solution

Assume a solution form

Consider a solution of the form $y=e^{rt}$. This is a common method to solve linear differential equations with constant coefficients.

Compute the derivatives

Compute the first and second derivatives of the assumed solution. $y=e^{rt}$ $y^{\prime }=r{e}^{rt}$ $y^{″}=r^2{e}^{rt}$

Substitute into the original equation

Substitute $y$, $y^{\prime }$, and $y^{″}$ back into the original differential equation: $4x{r}^2{e}^{rt}+2r{e}^{rt}+e^{rt}=0$

Factor out the exponential term

Factor out $e^{rt}$: $e^{rt}(4x{r}^2+2r+1)=0$

Set the characteristic equation

For the product to be zero, the term in the parentheses must be zero: $4x{r}^2+2r+1=0$

Solve the characteristic equation

Solve the quadratic equation for $r$: Use the quadratic formula, $r=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$, where $a=4x$, $b=2$, and $c=1$.

Use the quadratic formula

Using the quadratic formula, we get: $r=\frac{-2\pm \sqrt{4-16x}}{8x}$

Write the general solution

After solving for $r$, the general solution for the differential equation will be: $y(x)=C_1{e}^{r_{1}x}+C_2{e}^{r_{2}x}$, where $r_1$ and $r_2$ are the roots found in the previous step.

Key concepts

Characteristic Equation

To solve a second-order linear differential equation with constant coefficients, we start by forming the characteristic equation. This involves assuming a solution of the form $y=e^{rt}$.
Compute the first and second derivatives:

• $y^{\prime }=r{e}^{rt}$
• $y^{″}=r^2{e}^{rt}$

Next, substitute $y$, $y^{\prime }$, and $y^{″}$ back into the original differential equation:
In this exercise: $4x{r}^2{e}^{rt}+2r{e}^{rt}+e^{rt}=0$.
Factor out the exponential term $e^{rt}$: $e^{rt}(4x{r}^2+2r+1)=0$.
Since the exponential function $e^{rt}$ never equals zero for any real value of $r$, the term in the parentheses must be zero.
This gives us the characteristic equation: $4x{r}^2+2r+1=0$.

Exponential Solutions

With the characteristic equation formed, we can find exponential solutions. Exponential solutions are assumed solutions of the form $y=e^{rt}$, where $r$ is determined by solving the characteristic equation. The exponential function is very helpful in solving differential equations with constant coefficients because its derivatives are proportional to itself.
The roots $r$ we find from solving the characteristic equation indicate the solutions of the original differential equation:

• If roots are real and distinct, solutions take the form $y=C_1{e}^{r_{1}x}+C_2{e}^{r_{2}x}$.
• If roots are repeated, solutions involve terms like $y=(C_1+C_{2}x)e^{rx}$.
• If roots are complex, solutions include sine and cosine terms.

Quadratic Formula

To solve the characteristic equation, we often need the quadratic formula. It is used when the characteristic equation is in the form $a{r}^2+br+c=0$.
The quadratic formula is: \ $$r=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$ \
In our case, the characteristic equation is $4x{r}^2+2r+1=0$, so we identify:

• $a=4x$
• $b=2$
• $c=1$

Plug these into the quadratic formula: \
$$r=\frac{-2\pm \sqrt{2^2-4(4x)(1)}}{2(4x)}=\frac{-2\pm \sqrt{4-16x}}{8x}$$.
This yields the roots, which are the solutions $r$ for our differential equation.

General Solution

The general solution to a second-order linear differential equation combines the individual solutions corresponding to the roots of the characteristic equation. Once we determine the roots using the quadratic formula, we express the general solution.
Suppose the roots are $r_1$ and $r_2$, then the general solution is:
\

• If $r_1$ and $r_2$ are real, distinct roots: $y(x)=C_1{e}^{r_{1}x}+C_2{e}^{r_{2}x}$.
• If $r_1$ and $r_2$ are real, repeated roots: $y(x)=(C_1+C_{2}x)e^{rx}$.
• If the roots are complex conjugates: $r=\alpha \pm i\beta$, then the solution is: $y(x)=e^{\alpha x}(C_1\cos (\beta x)+C_2\sin (\beta x))$.

For the given exercise, the general solution would depend on the nature of the roots $r_1$ and $r_2$. The final general solution is: $y(x)=C_1{e}^{r_{1}x}+C_2{e}^{r_{2}x}$, where $r_1$ and $r_2$ are the roots calculated using the quadratic formula.