solve double differentiation with two limits

%% hello I am trying to solve the equation with double differentiation of M_z which is equal to w_x get the value of M_z(x) but could not solve it. Could you please help me to solve this equation. M_z(x= 0) = -150 and M_z(x= L) = 250 are given

clc;

close;

clear;

syms d_F ;

syms d_F_c;

syms dM_B;

%syms Theta_B;

syms x real;

syms u_mid;

syms u_c;

syms u_max;

syms a;

syms w_x;

syms M_z;

%syms n;

%Q = 2.5; %kip/ft

L = 30.; % ft

%a = 18; %assume maximum deflection at 15', 20'

E = 29000.*12^2; %ksi to kips/ft2

I = 1890./(12)^4; %ft^4

E_I = E.*I; %kip*ft^2;

%% deflection at mid of AB

%P = 1;

%R = 1;

%syms u_2;

%R = 1

w_x = @(x) -1*x/(L/3)*heaviside(x-L/3)*(1-heaviside(x-2*L/3));

w_x_1 = w_x(x);

eq1 = gradient(M_z(x),2) == -w_z(x);

soln1 = dsolve(eq1, M_z(0)==-150, M_z(L) == 250);

fplot(x, w_x(x), [0,L]);

Answer 1

[VBBV]

clc;
close;
clear;
syms d_F ;
syms d_F_c;
syms dM_B;
%syms Theta_B;
syms x real;
syms u_mid;
syms u_c;
syms u_max;
syms a;
% syms w_x;
syms M_z(x);
%syms n;
%Q = 2.5; %kip/ft
L  = 30.; % ft 
%a = 18;  %assume maximum deflection at 15', 20' 
E = 29000.*12^2; %ksi to kips/ft2
I = 1890./(12)^4; %ft^4
E_I = E.*I; %kip*ft^2;
%% deflection at mid of AB
%P = 1;
%R = 1;
%syms u_2;
%R = 1
w_x = @(x) -1*x/(L/3)*heaviside(x-L/3)*(1-heaviside(x-2*L/3));
w_x_1 = w_x(x)

w_x_1 = 

eq1 = diff(M_z(x),2) == -w_x_1

eq1 = 

soln1 = dsolve(eq1, [M_z(0)==-150, M_z(L) == 250])

soln1 = 

fplot(x, w_x(x), [0,L]); axis([0 30 -5 2])

[Milan]

Hello, It works but here I need to calculate M_z(x) which is moment. I don't think sol1 is M_z(x), is it?

[VBBV]

sol1 is for moment variable M_z(x) . please open a new question for your problem

[Milan]

I did it, could you please have a look?