Let's solve each part step-by-step for the ODE \( t x' - 4x = 0 \): --- ### **a) Is the ODE linear?** **Yes.** The ODE can be written as \( t x' - 4x = 0 \), or \( t x' = 4x \). This is linear in \( x \) and \( x' \) because both terms are first degree in \( x \) and \( x' \) and there are no products or nonlinear functions of \( x \) or \( x' \). --- ### **b) Is the ODE homogeneous?** **Yes.** The ODE is homogeneous because the right side is zero. There are no terms that are functions of \( t \) alone. --- ### **c) Find a solution by substituting \( x(t) = t^r \) and solving for \( r \).** Let \( x(t) = t^r \Rightarrow x'(t) = r t^{r-1} \). Substitute into the ODE: \[ t x' - 4x = 0 \implies t \cdot (r t^{r-1}) - 4 t^r = 0 \implies r t^r - 4 t^r = 0 \] \[ (r - 4) t^r = 0 \] Thus, \( r = 4 \). **So, \( x(t) = t^4 \) is a solution.** --- ### **d) Can you apply the superposition principle? If so, what other functions are also solutions?** **Yes,** the equation is linear and homogeneous, so the superposition principle applies: If \( x_1(t) \) and \( x_2(t) \) are solutions, then so is \( a x_1(t) + b x_2(t) \) for constants \( a, b \). From part (c), \( x(t) = t^4 \) is a solution. Any constant multiple \( c t^4 \) is also a solution. **All functions of the form \( x(t) = c t^4 \) are solutions.** --- ### **e) Solve by separation of variables and compare solutions.** Given: \[ t x' = 4x \implies \frac{x'}{x} = \frac{4}{t} \] Integrate both sides: \[ \int \frac{x'}{x} dt = \int \frac{4}{t} dt \implies \ln|x| = 4\ln|t| + C \implies |x| = e^C |t|^4 \implies x(t) = C_1 t^4 \quad (C_1 = \pm e^C) \] This matches the solution from parts (c) and (d). --- ### **f) Solutions for initial condition \( x(0) = 0 \). Are there infinitely many?** - **Constant solution:** \( x(t) = 0 \) satisfies \( x(0) = 0 \). - **Non-constant solution:** \( x(t) = c t^4 \); for \( x(0) = 0 \), any \( c \) works, but at \( t=0 \), \( x(0) = 0 \) for all \( c \). But \( t^4 \) is only defined for \( t \geq 0 \) (if we restrict to real, nonnegative \( t \)), or for all \( t \) if we allow real powers of negative numbers. However, as a function, \( x(t) = c t^4 \) for any \( c \) satisfies \( x(0) = 0 \). Thus, **there are infinitely many solutions** satisfying \( x(0) = 0 \): any \( x(t) = c t^4 \) with \( c \in \mathbb{R} \). --- ### **g) Solution for initial condition \( x(2) = 48 \).** From above, \( x(t) = c t^4 \) and \( x(2) = 48 \): \[ x(2) = c \cdot 2^4 = 16c = 48 \implies c = 3 \] **So, the solution is \( x(t) = 3 t^4 \).** --- ### **h) Restriction to \( t > 0 \); uniqueness** Let \( y(t) = t^{-r} x(t) \), where \( x \) is any solution and \( r = 4 \). So, \( y(t) = t^{-4} x(t) \). From above, all solutions are \( x(t) = c t^4 \), so \( y(t) = c \), a constant. If \( x \) also satisfies \( x(2) = 48 \), then \( y(2) = 2^{-4} x(2) = (1/16) \cdot 48 = 3 \), so \( c = 3 \). Thus, **if restricted to \( t > 0 \), the only solution satisfying \( x(2) = 48 \) is \( x(t) = 3 t^4 \)**. This is in contrast to part (f), where for \( t = 0 \) you could have infinitely many solutions. --- ### **i) Piecewise solution and infinite solutions for \( x(2) = 48 \).** Let \( x(t) = \begin{cases} c t^4, & t > 0 \\ 0, & t = 0 \\ b t^4, & t < 0 \end{cases} \) - For \( t > 0 \): \( x'(t) = 4 c t^3 \), so \( t x'(t) - 4x(t) = t (4c t^3) - 4(c t^4) = 4c t^4 - 4c t^4 = 0 \). - For \( t < 0 \): \( x'(t) = 4 b t^3 \), so \( t x'(t) - 4x(t) = t (4b t^3) - 4(b t^4) = 4b t^4 - 4b t^4 = 0 \). - For \( t = 0 \): \( x(0) = 0 \). \( x(t) \) is differentiable everywhere (as derivatives from left and right at \( t = 0 \) are both 0). To solve \( x(2) = 48 \): \[ x(2) = c \cdot 2^4 = 16c = 48 \implies c = 3 \] So for any real \( b \), setting \( c = 3 \), \( x(t) \) satisfies the ODE and \( x(2) = 48 \). **Thus, there are infinitely many solutions to the ODE with \( x(2) = 48 \), parameterized by \( b \).** --- ## **Summary Table** | Part | Answer | |------|--------| | a | Yes, it's linear | | b | Yes, it's homogeneous | | c | \( x(t) = t^4 \) | | d | Yes; all \( x(t) = c t^4 \) are solutions | | e | Separation gives \( x(t) = C_1 t^4 \), matching above | | f | Infinitely many solutions for \( x(0) = 0 \) (all \( x(t) = c t^4 \)) | | g | \( x(t) = 3 t^4 \) for \( x(2) = 48 \) | | h | Restricting \( t>0 \), only \( x(t) = 3 t^4 \) satisfies \( x(2) = 48 \) | | i | The piecewise function with \( c = 3 \) and arbitrary \( b \) is everywhere differentiable and solves the ODE, so there are infinitely many solutions to \( x(2) = 48 \) |