Question Statement

Show that $y_1=x^2$ and $y_2=x^3$ are both solutions of the differential equation: $x^2{y}^{\prime \prime }-4x{y}^{\prime }+6y=0$

(i) Are $c_1{y}_1$ and $c_2{y}_2$ also solutions? (ii) Check if $y_1+y_2$ is also a solution.

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Background and Explanation

To verify if a function is a solution to a differential equation, we calculate its derivatives and substitute them into the equation to see if it holds true. This problem also explores the Principle of Superposition, which states that for linear homogeneous differential equations, any linear combination of solutions is also a solution.

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Solution

The given differential equation is: $x^2{y}^{\prime \prime }-4x{y}^{\prime }+6y=0$

Verification of $y_1$ and $y_2$

Case I: $y_1=x^2$ First, we find the first and second derivatives: $y_1^{\prime }=2x$ $y_1^{\prime \prime }=2$

Now, we substitute $y_1$, $y_1^{\prime }$, and $y_1^{\prime \prime }$ into equation (1): $x^2(2)-4x(2x)+6(x^2)=0$ $2x^2-8x^2+6x^2=0$ $0=0$ Since the equation holds true, $y_1=x^2$ is a solution.

Case II: $y_2=x^3$ First, we find the derivatives: $y_2^{\prime }=3x^2$ $y_2^{\prime \prime }=6x$

Substitute $y_2$, $y_2^{\prime }$, and $y_2^{\prime \prime }$ into equation (1): $x^2(6x)-4x\left(3x^2\right)+6\left(x^3\right)=0$ $6x^3-12x^3+6x^3=0$ $0=0$ Since the equation holds true, $y_2=x^3$ is a solution.

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Part (i)

Checking $y=c_1{y}_1$: Let $y=c_1{x}^2$. The derivatives are: $y^{\prime }=2c_{1}x$ $y^{\prime \prime }=2c_1$

Substituting these into equation (1): $x^2\left(2c_1\right)-4x\left(2c_{1}x\right)+6(c_1{x}^2)=0$ $2c_1{x}^2-8c_1{x}^2+6c_1{x}^2=0$ $0=0$ Thus, $y=c_1{y}_1$ is also a solution.

Checking $y=c_2{y}_2$: Let $y=c_2{x}^3$. The derivatives are: $y^{\prime }=3c_2{x}^2$ $y^{\prime \prime }=6c_{2}x$

Substituting these into equation (1): $x^2\left(6c_{2}x\right)-4x\left(3c_2{x}^2\right)+6\left(c_2{x}^3\right)=0$ $6c_2{x}^3-12c_2{x}^3+6c_2{x}^3=0$ $0=0$ Thus, $y=c_2{y}_2$ is also a solution.

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Part (ii)

Checking $y=y_1+y_2$: Let $y=x^2+x^3$. We find the derivatives: $y^{\prime }=2x+3x^2$ $y^{\prime \prime }=2+6x$

Substituting $y$, $y^{\prime }$, and $y^{\prime \prime }$ into equation (1): $x^2(2+6x)-4x\left(2x+3x^2\right)+6\left(x^2+x^3\right)=0$ Distribute the terms: $2x^2+6x^3-8x^2-12x^3+6x^2+6x^3=0$ Group the $x^2$ and $x^3$ terms together: $(2x^2-8x^2+6x^2)+(6x^3-12x^3+6x^3)=0$ $0+0=0$ $0=0$ Hence, $y=y_1+y_2$ is also a solution of equation (1).

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Key Formulas or Methods Used

• Power Rule for Differentiation: $\frac{d}{dx}(x^n)=n{x}^{n-1}$
• Verification by Substitution: Plugging a function and its derivatives into an ODE to check for equality.
• Principle of Superposition: For a linear homogeneous equation, if $y_1$ and $y_2$ are solutions, then $c_1{y}_1+c_2{y}_2$ is also a solution.

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Summary of Steps

1. Calculate the first and second derivatives for $y_1=x^2$ and $y_2=x^3$.
2. Substitute these into the differential equation to confirm they satisfy it.
3. Multiply each solution by a constant ($c_1,c_2$) and repeat the substitution to show they remain solutions.
4. Add the two solutions together ($y_1+y_2$), find the new derivatives, and substitute them into the equation to verify the sum is also a solution.