Question

Verify that the given functions \(y_{1}\) and \(y_{2}\) satisfy the corresponding homogeneous equation; then find a particular solution of the given nonhomogeneous equation. In Problems 19 and \(20 g\) is an arbitrary continuous function. $$ t y^{\prime \prime}-(1+t) y^{\prime}+y=t^{2} e^{2 i}, \quad t>0 ; \quad y_{1}(t)=1+t, \quad y_{2}(t)=e^{t} $$

Short answer

Question: Verify that \(y_1(t) = 1 + t\) and \(y_2(t) = e^t\) satisfy the homogeneous equation \(t y^{\prime \prime}-(1+t) y^{\prime}+y=0\), and find a particular solution of the nonhomogeneous equation \(t y^{\prime \prime}-(1+t) y^{'}+y=t^2 e^{2i}\). Answer: Both \(y_1(t) = 1 + t\) and \(y_2(t) = e^t\) satisfy the homogeneous equation. The particular solution of the nonhomogeneous equation is \(y_p(t) = t^2 e^{2i}\). The general solution is given by \(y(t) = C_1(1 + t) + C_2 e^t + t^2 e^{2i}\).

Step-by-step solution

Verify that \(y_1\) and \(y_2\) satisfy the homogeneous equation

To do this, we need to plug in \(y_1\) and \(y_2\) into the homogeneous equation and check if they hold true. Homogeneous equation is: \(t y^{\prime \prime}-(1+t) y^{\prime}+y=0\) For \(y_1(t) = 1 + t\): The first derivative is: \(y^{\prime}_{1}(t) = 1\) The second derivative is: \(y^{\prime}_{2}(t) = 0\) Plug them into the homogeneous equation: \(t(0) - (1+t)(1) + (1+t) = 0\) This is indeed true. Therefore, \(y_1(t)\) satisfies the homogeneous equation. For \(y_2(t) = e^t\): The first derivative is: \(y^{\prime}_{1}(t) = e^t\) The second derivative is: \(y^{\prime}_{2}(t) = e^t\) Plug them into the homogeneous equation: \(t(e^t) - (1+t)(e^t) + e^t = 0\) Factor out \(e^t\): \(e^t (t - t - 1 + 1) = 0\) This is also true. Therefore, \(y_2(t)\) satisfies the homogeneous equation.

Find a particular solution using undetermined coefficients

We have a nonhomogeneous function in the form of \(t^2e^{2i}\). For this, we'll guess a particular solution of the form: \(y_p(t) = A t^2 e^{2i}\) First derivative of \(y_p(t)\): \(y_p^{\prime}(t) = 2 A t e^{2i} + 4 i A t^{2} e^{2i} \) Second derivative of \(y_p(t)\): \(y_p^{\prime\prime}(t) = 2 A e^{2i} + 8 i A t e^{2i} - 8 A t^{2} e^{2i}\) Plug \(y_p(t)\), \(y_p^{\prime}(t)\) and \(y_p^{\prime\prime}(t)\) into the nonhomogeneous equation: \(t(2Ae^{2i} + 8iAt e^{2i} - 8 At^2 e^{2i}) - (1 + t)(2A te^{2i} + 4 i At^2 e^{2i}) + At^2 e^{2i} = t^2 e^{2i}\) Now, we equate coefficients to match the nonhomogeneous term: \(A = 1\) (since this matches the coefficient of \(t^2e^{2i}\)) So now we have our particular solution: \(y_p(t) = t^2 e^{2i}\) Now we can write the general solution as a linear combination of the two homogeneous solutions and the particular solution: \(y(t) = C_1(1 + t) + C_2 e^t + t^2 e^{2i}\)

Key concepts

Understanding Homogeneous Equations

A homogeneous equation can be thought of as a puzzle where all the pieces fit perfectly to make zero. Officially, it's an ordinary differential equation that equals zero when all terms involve the function or its derivatives. Imagine it like a perfect balance - no matter how much you shake it, it always levels out to zero.

In the case of our exercise, consider the equation \(t y^{\'\'}-(1+t) y^{\'})+y=0\). To check if our piece, say \(y_1(t) = 1+t\), fits in this puzzle, we just need to replace \(y\) with \(1+t\) and see if it still balances. We calculate the first and second derivatives of \(y_1(t)\) and plug them into the equation. If we end up with zero, voila! We've found a piece that belongs. The same goes for our second piece, \(y_2(t) = e^t\), and if both satisfy the homogeneous equation, then we know these are good pieces of the puzzle that can help us solve more complicated ones, like our nonhomogeneous differential equation.

The Particular Solution

When you have a nonhomogeneous equation, think of it as a puzzle where pieces nearly fit, but there's always something sticking out. This 'extra stuff' is your nonhomogeneous term. The particular solution is like custom-carving a puzzle piece to perfectly cover that extra bit. It doesn't need to work with the entire puzzle—just this special part.

In our exercise, the nonhomogeneous term is \(t^2 e^{2i}\), so we're looking for a tailor-made piece, a particular solution that deals specifically with this term. We guess this special piece follows a similar pattern, which is why we choose it as \(At^2 e^{2i}\). After some fancy math-fu—differentiating and plugging it into the nonhomogeneous equation—we adjust the size of our piece, finding the value of \(A\) that makes it fit just right. Once that's done, we've effectively filled in the missing piece of our puzzle, creating the perfect particular solution for our problem!

The Method of Undetermined Coefficients

The method of undetermined coefficients is like a detective game where you guess who the culprit is and then look for clues to prove it. In mathematical terms, we make an educated guess about the form of a particular solution to our nonhomogeneous equation. Our 'culprits' might be based on the type of nonhomogeneous term we have. We then use our detective skills to find the 'dream team' of undetermined coefficients that make our guess solve the equation.

In the problem at hand, we're dealing with a nonhomogeneous term that resembles \(t^2 e^{2i}\), leading us to guess a similar form for our particular solution. We then use our investigative powers to differentiate our guess, plug the result into the equation, and compare coefficients to determine the exact values we need for our solution to be correct. The thrill of this method lies in finding precise values for our undetermined coefficients that make our initial guess into a solid, undeniable fact of the equation.