# 31.6 Solve the initial value problem

• MHB
• karush
In summary: A$In summary, the matrix equation y_1'= 2y_1+ y_2+ e^x can be solved for y_1(x) and y_2(x) by differentiating the equations and replacing y_2 by -y_1+ 2y_2. karush Gold Member MHB$\tiny{31.6}$Solve the initial value problem$Y'=\left|\begin{array}{rr}2 & 1 \\-1 & 2 \end{array}\right|Y
+\left|\begin{array}{rr}e^x \\0 \end{array}\right|,
\quad Y(0)=\left|\begin{array}{rr} 1 \\1 \end{array}\right| $ok so we have the form$y'=AY+G$rewrite as $$\displaystyle \left|\begin{array}{rr}y_1^\prime \\y_2^\prime \end{array}\right| =\left|\begin{array}{rr}2 & 1 \\-1 & 2 \end{array}\right| \left|\begin{array}{rr}y_1 \\y_2\end{array}\right| +\left|\begin{array}{rr}e^x \\0 \end{array}\right|$$ ok so the next thing to do is find eigenvalues of A so$\left| \begin{array}{cc}
-\lambda+2&1\\-1&-\lambda+2\end{array}
\right|
=\left(-\lambda+2\right)^{2}+1$so roots are$\lambda_{1}=2 + i, \qquad \lambda_{2}=2 - i$so far ? hopefully Yes, that is correct. Now, having found the eigenvalues, you need to find the corresponding eigenvectors. Let's look at solving this with a slightly different method. The matrix equation is equivalent to the pair of equations $$y_1'= 2y_1+ y_2+ e^x$$ and $$y_2'= -y_1+ 2y_2$$. Differentiate the first equation again to get $$y_1''= 2y_1'+ y_2'+ e^x$$. Replace that $$y_2'$$ by $$-y_1+ 2y_2$$ from the second equation: $$y_1''= 2y_1'- y_1+ 2y_2+ e^x$$. From the first equation, $$y_2= y_1'- 2y_1- e^x$$ so $$y_1''= 2y_1'- y_1+ 2(y_1'- 2y_1- e^x)+ e^x$$ or $$y_1''- 4y_1'+ 5y_1= - e^x$$. The associated homogeneous equation is y_1''- 4y_1'+ 5y_1= 0 which has characteristic equation $$r^2- 4r+ 5= 0$$. By the quadratic formula, that has roots $$\frac{4\pm\sqrt{16- 20}}{2}= 2\pm i$$ just as you say. That tells us that the general solution to the associated homogeneous equation is $$y_1(x)= e^{2x}(C_1cos(x)+ C_2sin(x))$$. We look for a solution to the entire equation of the form $$y_1(x)= Ae^x$$. Then $$y'(x)= y''(x)= Ae^x$$ and the equation becomes $$Ae^x- 4Ae^x+ 5Ae^x= 2Ae^x= -e^x$$ so $$A= -\frac{1}{2}. We have [tex]y_1(x)= e^{2x}(C_1cos(x)+ C_2sin(x))- \frac{1}{2}e^x$$. Solve for $$y_2(x)$$ from $$y_2= y_1'- 2y_1- e^x$$. (It is odd that you posted this, which involves a differential equation, under "Linear and Abstract Algebra" while you posted [FONT=Tahoma,Calibri,Verdana,Geneva,sans-serif]"https://mathhelpboards.com/differential-equations-17/14-1-find-vector-v-will-satisfy-system-26203.html", [/FONT] which does not, under "differential equations"!) Last edited by a moderator: HallsofIvy said: (It is odd that you posted this, which involves a differential equation, under "Linear and Abstract Algebra" while you posted "https://mathhelpboards.com/differential-equations-17/14-1-find-vector-v-will-satisfy-system-26203.html", which does not, under "differential equations"!) https://www.physicsforums.com/attachments/9086 will they combined the DE with LA so I often label it wrong poor reason I know. Just hope I will be ready for the next class, whatever commonly comes after this one the last week they started the chapter "First Order Ordinary Differential Equations" Which I posted some of earlier on MHB ok well back in the sattle solve$\displaystyle y_2= y_1'- 2y_1- e^x$rewrite as$\displaystyle y_2-e^x=y_1^\prime-\frac{1}{2}y_1\$ok assume the next step is the factor

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## 1. What is an initial value problem?

An initial value problem is a type of differential equation that involves finding a function that satisfies both a given differential equation and a set of initial conditions. These initial conditions specify the value of the function at a particular point or points.

## 2. How do you solve an initial value problem?

To solve an initial value problem, you need to first find the general solution of the given differential equation. Then, you can use the initial conditions to determine the particular solution that satisfies both the differential equation and the initial conditions.

## 3. What is the significance of solving an initial value problem?

Solving an initial value problem allows us to find a specific function that satisfies both the differential equation and the given initial conditions. This can help us make predictions about the behavior of a system or model real-world phenomena.

## 4. Can an initial value problem have multiple solutions?

Yes, an initial value problem can have multiple solutions. This is because the general solution of a differential equation can have arbitrary constants that can be determined using the initial conditions. Therefore, there can be multiple particular solutions that satisfy the initial value problem.

## 5. What are some real-world applications of solving initial value problems?

Initial value problems have various applications in fields such as physics, engineering, and economics. They can be used to model and predict the behavior of systems such as population growth, chemical reactions, and electrical circuits. They also play a crucial role in solving differential equations in mathematical finance and control theory.

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