The image provides a differential equation problem with initial conditions. The problem is a third-order linear homogeneous differential equation with the following form:\[ y''' - 4y'' + 13y' + 50y = 0 \]Initial conditions are given as:- \( y(0) = 4 \)- \( y'(0) = 10 \)- \( y''(0) = 42 \)The task is to solve for \( y(t) \), and there is a box meant for the expression of \( y(t) \). **Problem Statement:**Solve the differential equation:\[ y^{(4)} + 18y'' + 81y = 0 \]with the initial conditions:\[ y(0) = -4, \quad y'(0) = 8, \quad y''(0) = 54, \quad y'''(0) = 0 \]**Explanation:**This is a fourth-order linear homogeneous differential equation with constant coefficients. The task is to find a function \( y(t) \) that satisfies this equation, given the specified initial conditions. The solution requires finding the characteristic equation, solving for the roots, and then constructing the general solution. The initial conditions are used to find specific values for the constants in the general solution.There are no graphs or diagrams included in this problem.

Given: The differential equation is y'''-4y''+13y'+50y=0 with initial conditions y0=4, y'0=10, y''0=42. To find: Solution yt of the differential equation y'''-4y''+13y'+50y=0. Concepts used: 1. The general solution of the differential equation with real and distinct roots is defined by yt=c1em1t+c2em2t+...+cnemnt. 2. The general solution of the differential equation with imaginary roots α±iβ is defined by yt=eαtc1cosβt+c2sinβt.The differential equation is y'''-4y''+13y'+50y=0. The characteristic equation is D3-4D2+13D+50y=0. The auxiliary equation is ⇒ D3-4D2+13D+50=0⇒D3+2D2-6D2-12D+25D+50=0⇒D2D+2-6DD+2+25D+2=0⇒ D+2D2-6D+25=0 ⇒Either D+2=0 or D2-6D+25=0. Case 1: If D+2=0, then ⇒D=-2 Case 2: If D2-6D+25=0, then ⇒D=--6±-62-412521⇒D=6±36-1002⇒D=6±-642⇒D=6±64i22⇒D=6±8i2∴ D=3±4i Therefore, the roots of the differential equation are D=-2, D=3+4i, and D=3-4i. Hence, the general solution of the differential equation is yt=c1e-2t+e3tc2cos4t+c3sin4t, where c1, c2, c3 are arbitrary constants to be determined.Differentiate the solution yt=c1e-2t+e3tc2cos4t+c3sin4t with respect to t. y't=ddtc1e-2t+e3tc2cos4t+c3sin4t =c1ddte-2t+e3tddtc2cos4t+c3sin4t+c2cos4t+c3sin4tddte3t =c1-2e-2t+e3tc2-4sin4t+c34cos4t+c2cos4t+c3sin4t3e3t =-2c1e-2t-4c2e3tsin4t+4c3e3tcos4t+3c2e3tcos4t+3c3e3tsin4t =-2c1e-2t+-4c2+3c3e3tsin4t+4c3+3c2e3tcos4t ...1 Differentiate equation (1) with respect to t. y''t=ddt-2c1e-2t+-4c2+3c3e3tsin4t+4c3+3c2e3tcos4t =-2c1ddte-2t+-4c2+3c3ddte3tsin4t+4c3+3c2ddte3tcos4t =-2c1-2e-2t+-4c2+3c3sin4tddte3t+e3tddtsin4t+4c3+3c2cos4tddte3t+e3tddtcos4t =4c1e-2t+-4c2+3c3sin4t3e3t+e3t4cos4t+4c3+3c2cos4t3e3t+e3t-sin4t =4c1e-2t+-4c2+3c33e3tsin4t+4e3tcos4t+4c3+3c23e3tcos4t-e3tsin4t ...2Determine the constants. The initial conditions are y0=4, y'0=10, y''0=42. Case 1: At y0=4. Substitute 0 for t and 4 for y0 in yt=c1e-2t+e3tc2cos4t+c3sin4t. ⇒y0=c1e-20+e30c2cos40+c3sin40⇒ 4=c1e0+e0c2cos0+c3sin0⇒ 4=c11+1c21+c30∴ 4=c1+c2 ...3 Case 2: At y'0=10. Substitute 0 for t and 10 for y'0 in equation (1). ⇒y'0=-2c1e-20+-4c2+3c3e30sin40+4c3+3c2e30cos40⇒ 10=-2c1e0+-4c2+3c3e0sin0+4c3+3c2e0cos0⇒ 10=-2c11+-4c2+3c310+4c3+3c211∴ 10=-2c1+4c3+3c2 ...4 Case 3: At y''0=42. Substitute 0 for t and 42 for y''0 in equation (2). ⇒y''0=4c1e-20+-4c2+3c33e30sin40+4e30cos40+4c3+3c23e30cos40-e30sin40⇒ 42=4c1e0+-4c2+3c33e0sin0+4e0cos0+4c3+3c23e0cos0-e0sin0⇒ 42=4c11+-4c2+3c3310+411+4c3+3c2311-10⇒ 42=4c1+-4c2+3c30+4+4c3+3c23-0⇒ 42=4c1+-4c2+3c34+4c3+3c23⇒ 42=4c1-16c2+12c3+12c3+9c2⇒ 42=4c1-7c2+24c3 …5 Multiply equation (4) by 6 and subtract equation (5) from the result. ⇒6-2c1+4c3+3c2-4c1-7c2+24c3=610-42⇒ -12c1+24c3+18c2-4c1+7c2-24c3=60-42 ⇒ -16c1+25c2=18 …6 Multiply equation (3) by 16 and add with equation (6). ⇒16c1+c2+-16c1+25c2=164+18⇒ 16c1+16c2-16c1+25c2=64+18⇒ 41c2=82⇒ c2=8241∴ c2=2 Substitute 2 for c2 in equation (3). ⇒c1+2=4⇒ c1=4-2∴ c1=2 Substitute 2 for c1 and c2 in equation (4). ⇒-22+4c3+32=10⇒ -4+4c3+6=10⇒ 4c3=10-2⇒ 4c3=8⇒ c3=84∴ c3=2Now, substitute 2 for c1, c2, and c3 in equation (4). ∴yt=2e-2t+e3t2cos4t+2sin4t Hence, the solution of the differential equation y'''-4y''+13y'+50y=0 with initial conditions y0=4, y'0=10, y''0=42 is yt=2e-2t+e3t2cos4t+2sin4t.