Question

The initial-conditions $\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}{\mathit{y}}_{\mathbf{0}}\mathbf{,}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}{\mathit{y}}_{\mathbf{1}}$apply to each of the following differential equations:

$$\begin{gathered} {\mathit{x}}^{\mathbf{2}}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{=}\mathbf{0} \\ {\mathit{x}}^{\mathbf{2}}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathbf{2}\mathit{x}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{=}\mathbf{0} \\ {\mathit{x}}^{\mathbf{2}}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathbf{4}\mathit{x}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{+}\mathbf{6}\mathit{y}\mathbf{=}\mathbf{0} \end{gathered}$$

Short answer

For the first D.E:

$y_0=c_1$and$y_1=c_2$

For the second D.E:

$y_0=0$and$y_1=c_1$

For the third D.E:

$y_0=0$and$y_1=0$

Step-by-step solution

Given information

$$\begin{gathered} x^2{y}^{\text{'}\text{'}}-2x{y}^{\text{'}}+2y=0 \\ {x}^2{y}^{\text{'}\text{'}}-4x{y}^{\text{'}}+6y=0 \\ y\left(0\right)=y_0,y^{\text{'}}\left(0\right)=y_1 \end{gathered}$$

Find general solution

We have the three differential equations

$$\begin{gathered} x^2{y}^{\text{'}\text{'}}=0 \\ {x}^2{y}^{\text{'}\text{'}}-2x{y}^{\text{'}}+2y=0 \\ {x}^2{y}^{\text{'}\text{'}}-4x{y}^{\text{'}}+6y=0 \end{gathered}$$

With the initial conditions

$$\begin{gathered} y\left(0\right)=y_0 \\ {y}^{\text{'}}\left(0\right)=y_1 \end{gathered}$$

And we have to obtain the values y₀ and y₁for each differential equation that make it having solution as the following:

First, for the differential equation

$x^2{y}^{\text{'}\text{'}}=0....\left(1\right)$

$$\begin{gathered} y^{\text{'}}=m{x}^{(m-1)} \\ {y}^{\text{'}\text{'}}=m(m-1)x^{(m-2)}....\left(a\right) \end{gathered}$$

After that, substitute from equation (a) into equation (1), then we have

$$\begin{gathered} x^2\left[m(m-1)x^{(m-2)}\right]=0 \\ m(m-1)x^m=0 \end{gathered}$$

Since $x^m$cannot be equal 0, then we have the auxiliary solution as:

$m(m-1)=0$

Then we the roots as:

$m_1=0$and$m_2=1$

Then we can obtain the general solution for the first differential equation$x^{2}y\text{'}\text{'}=0$ as:

$y\left(x\right)=c_1+c_{2}x....\left(b\right)$

To apply the initial conditions

Now, we have to apply the initial condition$\left(x,y\right)=\left(0,y_0\right)$ into equation (b), and then we have;

$$\begin{gathered} y_0=c_1+0 \\ {y}_0=c_1 \end{gathered}$$

After that, differentiate equation (b) with respect to x , then we have

$y^{\text{'}}\left(x\right)=c_2$

Then apply the initial condition $\left(x,y\text{'}\right)=\left(0,y_1\right)$, and then we have

$y_1=c_2$

Then the first differential equations has the solution $y\left(x\right)=y_0+y_{1}x,$with $y_0=c_1$and$y_1=c_2$

Find first differential equation

Second, for the differential equation

$x^2{y}^{\text{'}\text{'}}-2x{y}^{\text{'}}+2y=0....\left(2\right)$

Assume that $y=x^m$ then differentiate with respect to, then we have

$$\begin{gathered} y^{\text{'}}=m{x}^{(m-1)}....\left(c\right) \\ {y}^{\text{'}\text{'}}=m(m-1)x^{(m-2)}....\left(d\right) \end{gathered}$$

After that substitute$y=x^m$ with and from the equations (c) and (d) into equation (2), then we have;

$$\begin{gathered} x^2\left[m(m-1)x^{(m-2)}\right]-2x\left[m{x}^{(m-1)}\right]+2x^m=0 \\ m(m-1)x^m-2m{x}^m+2x^m=0 \\ \left[m(m-1)x^m-2(m-1)x^m\right]=0 \\ (m-1)(m-2)x^m=0 \end{gathered}$$

Since $x^m$cannot be equal 0, then we have auxiliary solution as

$(m-1)(m-2)=0$

Then we have the roots as

$m_1=1$and$m_2=2$

Then we can obtain general solution for the first differential equation

$x^2{y}^{\text{'}\text{'}}-2x{y}^{\text{'}}+2y=0as$

$y\left(x\right)=c_{1}x+c_2{x}^2.....\left(e\right)$

Now, we have to apply the initial condition$(x,y)=\left(0,y_0\right)$ into equation e ,and then we have

$y{}_0=0$

After that, differentiate equation e with respect to x , then we have

$y^{\text{'}}\left(x\right)=c_1+2c_{2}x$

Then apply the initial condition $\left.(x,y^{\text{'}}\right)=\left(0,y_1\right)$, and then we have

$y_1=c_1$

Then the first differential equation has a solution with$y_0=0$ and$y_1=c_1$

Find first differential equation

Third, for the differential equation

$x^2{y}^{\text{'}\text{'}}-4x{y}^{\text{'}}+6y=0....\left(3\right)$

Assume that $y=x^m$ then differentiate with respect to x, then we have

$$\begin{gathered} y^{\text{'}}=m{x}^{(m-1)}....\left(f\right) \\ {y}^{\text{'}\text{'}}=m(m-1)x^{(m-2)}....\left(g\right) \end{gathered}$$

After that substitute with$y=x^m$and from the equations (f) and (g) into equation (3), then we have

$$\begin{gathered} x^2\left[m(m-1)x^{(m-2)}\right]-4x\left[m{x}^{(m-1)}\right]+6x^m=0 \\ m(m-1)x^m-4m{x}^m+6x^m=0 \\ \left(m^2-5m+6\right)x^m=0 \end{gathered}$$

Since $x^m$cannot be equal 0, then we have auxiliary solution as

$$\begin{gathered} m^2-5m+6=0 \\ (m-2)(m-3)=0 \end{gathered}$$

Then we have the roots as

localid="1664310509422" $m_1=2$andlocalid="1664310504269" $m_2=3m_2=3$

Then we can obtain solution for the first differential equation

$$\begin{gathered} x^2{y}^{\text{'}\text{'}}-2x{y}^{\text{'}}+2y=0as \\ y\left(x\right)=c_1{x}^2+c_2{x}^3.....\left(h\right) \end{gathered}$$

Now, we have to apply the initial condition $(x,y)=\left(0,y_0\right)$ into equation (h), and then we have:

$y{}_0=0$

After that, differentiate equation (h) with respect to x , then we have

$y^{\text{'}}\left(x\right)=2c_{1}x+3c_2{x}^2$

Then apply the initial condition $\left.(x,y^{\text{'}}\right)=\left(0,y_1\right)$, and then we have

$y_1=0$

Then the first differential equation has a solution with $y_0=0$and$y_1=0$

conclusion

For the first D.E:

$y_0=c_1$and$y_1=c_2$

For the second D.E:

$y_0=0$and$y_1=c_1$

For the third D.E:

$y_0=0$and $y_1=0$