Question

solve the second order differential equation d2y/dx2−4y=12x,y(0)=4,y′(0)=1

Answer 1

Answer on Question #69650 – Math – Differential Equations

Question

Solve the second order differential equation

$$\frac {\partial^ {2} y}{\partial x ^ {2}} - 4 y = 12x, \quad y(0) = 4, \quad y'(0) = 1.$$

Solution

First, let's solve the corresponding homogenous equation $y'' - 4y = 0$. It's linear, so we can solve it using characteristic equation:

$$\begin{array}{l} t^{2} - 4 = 0, \\ (t + 2)(t - 2) = 0, \\ t_{1} = -2, \quad t_{2} = 2 \end{array}$$

There are two distinct eigenvalues, hence the general solution of this homogenous equation is

$$y_{0}(x) = C_{1} e^{t_{1}x} + C_{2} e^{t_{2}x} = C_{1} e^{-2x} + C_{2} e^{2x}.$$

Second, let's find any particular solution of the original non-homogenous equation. Right-hand side of the initial non-homogeneous differential equation is a polynomial of the first order, $12x$, therefore, we'll look for a particular solution in polynomial form of the same order,

$$y_{1}(x) = ax + b.$$ $$y_{1} = ax + b, \quad y_{1}^{\prime \prime} = 0,$$ $$\begin{array}{l} y^{\prime \prime} - 4y = 0 - 4(ax + b) = -4ax - b = 12x, \\ \left\{ \begin{array}{l} -4a = 12 \\ -b = 0 \end{array} \right. \rightarrow \left\{ \begin{array}{l} a = -3 \\ b = 0. \end{array} \right. \end{array}$$

Thus, $y_{1}(x) = -3x$ is a particular solution of the non-homogenous equation. Altogether, the general solution of the non-homogenous equation is the sum of any particular solution and the general solution of the corresponding homogenous equation:

$$y(x) = y_{0}(x) + y_{1}(x) = C_{1} e^{-2x} + C_{2} e^{2x} - 3x.$$

Finally, let's apply initial conditions: $y(0) = 4$ and $y'(0) = 1$.

$$\begin{array}{l} y^{\prime} = \left(C_{1} e^{-2x} + C_{2} e^{2x} - 3x\right)^{\prime} = -2C_{1} e^{-2x} + 2C_{2} e^{2x} - 3, \\ y^{\prime}(0) = -2C_{1} e^{-2\cdot 0} + 2C_{2} e^{2\cdot 0} - 3 = -2C_{1} + 2C_{2} - 3, \\ y(0) = C_{1} e^{-2\cdot 0} + C_{2} e^{2\cdot 0} - 3\cdot 0 = C_{1} + C_{2}. \end{array}$$

To do this, we need to solve the system:

$$\left\{ \begin{array}{l} C_{1} + C_{2} = 4 \\ -2C_{1} + 2C_{2} - 3 = 1 \end{array} \right. \rightarrow \left\{ \begin{array}{l} C_{2} = 4 - C_{1} \\ C_{2} = C_{1} + 2 \end{array} \right. \rightarrow \left\{ \begin{array}{l} C_{1} + 2 = 4 - C_{1} \\ C_{2} = C_{1} + 2 \end{array} \right. \rightarrow \left\{ \begin{array}{l} C_{1} = 1 \\ C_{2} = C_{1} + 2 \end{array} \right. \rightarrow \left\{ \begin{array}{l} C_{1} = 1 \\ C_{2} = 3. \end{array} \right.$$

Thus, the solution of the initial value problem is $y(x) = e^{-2x} + 3e^{2x} - 3x$.

**Answer**: $y(x) = e^{-2x} + 3e^{2x} - 3x$.

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