The answer is ideal in your question. Integration is the straight-forward way to prove it.
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Assuming $y""(x) = 0$ as a differential equation over the genuine variable $x$ :
$$y""(x) = 0 Rightarrowhead int y""(x)dx =int0dxRightarrowhead y"(x) = A Rightarrowint y"(x)dx = int Adx Rightarrow$$
$$y(x) = Ax + B oom , area extwhere area A,B oom extconstants oom in lutz-heilmann.infobb R$$
If a function"s derivative is zero, then the function is constant; this is a crucial allude regarding integration. So you know that $y"(x)=A$ for some consistent $A$. Now consider $z(x)=Ax$. Then $$(z-y)"=z"-y"=A-A=0. $$So $y(x)-z(x)=B$ for some consistent $B$. That is, $y(x)=Ax+B$.
In basic, for some nice sufficient differential equations one deserve to prove presence and uniqueness results. So, for instance, for a second order linear, homogeneous, continuous coeffective equation, one can prove that there is a distinct solution offered that the worth of the function and also its derivative are prescribed at a solitary point. Then, if you discover a solution to the initial value trouble, you will certainly recognize its the only one because of the theorem.
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