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Magazine Chapter 4: Problem 11
\(y=c_{1} t^{-1 / 2}+c_{2} t^{3}, t>0,2 t^{2} y^{\prime \prime}-3 t y^{\prime}-3 y=0\)
Short Answer
Expert verified
The function \( y = c_{1} t^{-1 / 2} + c_{2} t^{3} \) satisfies the differential equation.
Step by step solution
01
- Write the Given Differential Equation
Write down the differential equation to see what needs to be solved:\[2 t^{2} y^{\text{''}} - 3 t y^{'} - 3 y = 0\]
02
- Identify and Double Check the Proposed Solution
Recognize the given solution and substitute it into the differential equation:\[ y = c_{1} t^{-1 / 2} + c_{2} t^{3} \]Ensure that this function is eligible by checking continuity and differentiability.
03
- Calculate the First Derivative
Find the first derivative of the proposed solution:\[ y^{'} = -\frac{1}{2} c_{1} t^{-3/2} + 3 c_{2} t^{2}\]
04
- Calculate the Second Derivative
Compute the second derivative of the proposed solution:\[ y^{''} = \frac{3}{4} c_{1} t^{-5/2} + 6 c_{2} t\]
05
- Plug Derivatives into the Differential Equation
Substitute \( y \), \( y^{'} \), and \( y^{''} \) back into the differential equation:\[2 t^{2} \left( \frac{3}{4} c_{1} t^{-5/2} + 6 c_{2} t \right) - 3 t \left( -\frac{1}{2} c_{1} t^{-3/2} + 3 c_{2} t^{2} \right) - 3 \left( c_{1} t^{-1/2} + c_{2} t^{3} \right) = 0\]
06
- Simplify the Equation
Simplify and combine like terms to verify the equality:\[ \left( \frac{3}{2} c_{1} t^{-1/2} + 12 c_{2} t^{3} \right) + \left( \frac{3}{2} c_{1} t^{-1/2} - 9 c_{2} t^{3} \right) - 3 c_{1} t^{-1/2} - 3 c_{2} t^{3} = 0\]Combine terms:\[ 0 = 0\]
07
- Confirm the Solution
Since the left-hand side equals the right-hand side of the equation, the function \( y = c_{1} t^{-1 / 2} + c_{2} t^{3} \) is a solution of the given differential equation.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
First Derivative
The first derivative of a function measures how the function's value changes as its input changes. For the given equation, recall that our proposed solution is \( y = c_{1} t^{-1 / 2} + c_{2} t^{3} \). To find the first derivative (\( y' \)), we apply the power rule of differentiation. The derivative of \( t^{n} \) is \( n t^{n-1} \). Applying this rule, we get:
\( y^{'} = -\frac{1}{2} c_{1} t^{-3/2} + 3 c_{2} t^{2} \).
This tells us how the function changes with respect to \( t \). Each term in the derivative corresponds to a term in the original function with its exponent reduced by one and multiplied by the original exponent.
\( y^{'} = -\frac{1}{2} c_{1} t^{-3/2} + 3 c_{2} t^{2} \).
This tells us how the function changes with respect to \( t \). Each term in the derivative corresponds to a term in the original function with its exponent reduced by one and multiplied by the original exponent.
Second Derivative
The second derivative indicates how the rate of change of a function is itself changing. It's essentially the derivative of the first derivative. For our function \( y \) with the first derivative \( y^{'} = -\frac{1}{2} c_{1} t^{-3/2} + 3 c_{2} t^{2} \), we differentiate again to find \( y^{''} \):
\( y^{''} = \frac{3}{4} c_{1} t^{-5/2} + 6 c_{2} t \).
Each term has been differentiated again using the power rule. The second derivative helps us understand the curvature of the function's graph and is crucial in solving higher-order differential equations.
\( y^{''} = \frac{3}{4} c_{1} t^{-5/2} + 6 c_{2} t \).
Each term has been differentiated again using the power rule. The second derivative helps us understand the curvature of the function's graph and is crucial in solving higher-order differential equations.
Verification of Solutions
To verify if a proposed function is a solution to a differential equation, we substitute the function and its derivatives back into the equation and check if it holds true. Here, we substitute \( y \), \( y^{'} \), and \( y^{''} \) into the given differential equation:
\( 2 t^{2} \left( \frac{3}{4} c_{1} t^{-5/2} + 6 c_{2} t \right) - 3 t \left( -\frac{1}{2} c_{1} t^{-3/2} + 3 c_{2} t^{2} \right) - 3 \left( c_{1} t^{-1/2} + c_{2} t^{3} \right) \) = 0.
Simplifying step-by-step, we combine like terms until we verify the left-hand side equals zero, proving that indeed \( y = c_{1} t^{-1 / 2} + c_{2} t^{3} \) satisfies the differential equation.
\( 2 t^{2} \left( \frac{3}{4} c_{1} t^{-5/2} + 6 c_{2} t \right) - 3 t \left( -\frac{1}{2} c_{1} t^{-3/2} + 3 c_{2} t^{2} \right) - 3 \left( c_{1} t^{-1/2} + c_{2} t^{3} \right) \) = 0.
Simplifying step-by-step, we combine like terms until we verify the left-hand side equals zero, proving that indeed \( y = c_{1} t^{-1 / 2} + c_{2} t^{3} \) satisfies the differential equation.
Homogeneous Differential Equation
A homogeneous differential equation is one where every term is a function of the dependent variable and its derivatives. Homogeneous equations can often be solved by finding characteristic equations or by recognizing patterns in the solutions. The given differential equation \( 2 t^{2} y^{''}-3 t y^{'} -3 y = 0 \) is homogeneous, meaning it equates to zero.
In this case, terms involving \( y \), \( y' \), and \( y^{''} \) are weighted by functions of \( t \). Such equations are often solved by proposing solutions in the form of power series or exponential functions. Here we've used powers of \( t \), specifically \( t^{-1/2} \) and \( t^{3} \) to fit the homogeneous equation.
In this case, terms involving \( y \), \( y' \), and \( y^{''} \) are weighted by functions of \( t \). Such equations are often solved by proposing solutions in the form of power series or exponential functions. Here we've used powers of \( t \), specifically \( t^{-1/2} \) and \( t^{3} \) to fit the homogeneous equation.
