Question

Use the substitution t = -x to solve the given initial-value problem on the interval$\mathbf{(}\mathbf{-}\mathbf{\infty }\mathbf{,}\mathbf{0}\mathbf{)}$.$\mathbf{4}{\mathit{x}}^{\mathbf{2}}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathit{y}\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{=}\mathbf{2}\mathbf{,}\mathit{y}\mathbf{\text{'}}\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{=}\mathbf{4}$

Short answer

The solution of the given initial value by substituting the given equation is $y=2(-x{)}^{\frac{1}{2}}+3(-x{)}^{\frac{1}{2}}\ln (-x),x<0$

Step-by-step solution

Deriving the given solution by Product Rule and Chain Rule

$y\text{'}$Using the substitution t = - x, it follows from the Product Rule and Chain Rule that

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{dy}{dt}\frac{dt}{dx}=-\frac{dy}{dt}\\ \frac{d^{2}y}{d{x}^2}& =& -\frac{d}{dx}\left(\frac{dy}{dt}\right)\\ & =& -\frac{d}{dt}\left(\frac{dy}{dt}\right)\frac{dt}{dx}\\ & =& \frac{d^{2}y}{d{t}^2}\end{array}$

Substituting in the given differential and simplifying yields

$4t^{2}y\text{'}\text{'}+y=0$

The term is missing in the given Cauchy-Euler equation; nevertheless, the substitution$y=t^m$ yields

$$\begin{gathered} 4t^{2}y\text{'}\text{'}+y=0 \\ 4t^2\left[m(m-1)t^{m-2}\right]+t^m=0 \\ 4m(m-1)t^m+t^m=0 \\ {t}^m\left(4m^2-4m+1\right)=0 \end{gathered}$$

When $4m^2-4m+1=0$from the quadratic formula we find that the roots are$m_1=m_2=\frac{1}{2}$ . Therefore, the general solution from repeated real roots is

$:\text{\#}4257\text{b}2;\text{y}=c_1{t}^{\frac{1}{2}}+c_2{t}^{\frac{1}{2}}\ln t$

Deriving the given solution by the differential equation

$y\left(1\right)=2$The derivative for the solution is,

$y\text{'}=\frac{1}{2}{c}_1{t}^{\frac{1}{2}}+c_2\left(\frac{1}{2}{t}^{\frac{1}{2}}\ln t+t^{\frac{1}{2}}\frac{1}{t}\right)$

$$\begin{gathered} =\frac{c_1}{2}{t}^{\frac{1}{2}}+c_2\left(\frac{1}{2}{t}^{\frac{1}{2}}\ln t+t^{\frac{1}{2}}\right) \\ =t^{-\frac{1}{2}}\left[\frac{c_1}{2}+c_2\left(\frac{\ln t}{2}+1\right)\right] \end{gathered}$$

Since$y\left(1\right)=2$ the initial conditions become$y\left(1\right)=2andy\text{'}\left(1\right)=4$ Substituting in the solution gives

$$\begin{gathered} c_1{\left(1\right)}^{\frac{1}{2}}+c_2{\left(1\right)}^{\frac{1}{2}}\ln 1=2 \\ {c}_1+c_2\left(0\right)=2 \\ {c}_1=2 \end{gathered}$$

Substituting$y\text{'}\left(1\right)=-4$ in the derivative equation gives

$$\begin{gathered} \frac{1}{2}\left[\frac{c_1}{2}+c_2\left(\frac{\ln 1}{2}+1\right)\right]=4 \\ \frac{c_1}{2}+c_2(0+1)=4 \end{gathered}$$

$$\begin{gathered} \frac{2}{2}+c_2=4c_1=2 \\ 1+c_2=4 \\ {c}_2=3 \end{gathered}$$

Hence, the general solution is $y=2t^{\frac{1}{2}}-+3t^{\frac{1}{2}}\ln t$and upon reconstituting $t=-x$gives the general solution:

$y=2(-x{)}^{\frac{1}{2}}+3(-x{)}^{\frac{1}{2}}\ln (-x),x<0$

Here x is negative because the logarithm of negative numbers is not defined.