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I believe I need to guess a general solution of $y$ for this boundary value problem. However I am not sure what that guess is. Can someone explain if I am on the right track or not? If so what is the guess for the general solution of y?

Jnyeboah93
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We try $y = ax+b$. Substitute into the original equation, we get $b=1$. It happens to fit the first boundary condition. The second boundary condition gives $a=2$. Since the boundary condition uniquely determines this equation, this above solution is the unique one.

For a more general setting $y''(x) = −xy' (x) + y(x) − \beta$, $\beta\ne 1$, the above trial solution ceases to work. But after differentiating the equation once, we have $$y'''=-xy''.$$ It is a first order ODE of $y''$. Solve, $$\frac{y''(x)}{y''(x_0)}=\exp\bigg(-\frac{x^2-x_0^2}{2}\bigg),$$ for some $x_0$ where $y''(x_0)\ne 0$. $x_0$ exists. Otherwise $y$ is reduced to the linear form that fails to satisfy the equation.

You can integrate and match the boundary condition to arrive at the solution.

Hans
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  • How do you get $y′′′=−xy′′$ when you differentiate the equation? – Jnyeboah93 May 20 '15 at 19:20
  • Have you tried differentiating the expressions on both sides of the equation? – Hans May 20 '15 at 19:27
  • yes. I don see how you get your result. Also i do not understand how it helps. Can you elaborate please? – Jnyeboah93 May 20 '15 at 19:32
  • What do you get differentiating the right hand side of the original equation? – Hans May 20 '15 at 19:47
  • I have added the (unique) solution for this particular equation in the first paragraph. You can skip the general solution from the second paragraph down, so the question of differentiating the equation becomes moot. – Hans May 20 '15 at 20:43
  • Thanks I understand your first paragraph. Am I right in saying that $y=ax +b$ is an educated guess for the form of the unique solution? If so what form does the differential equation have to be in for $y=ax +b$ to be a suitable educated guess? – Jnyeboah93 May 21 '15 at 09:26
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    Yes, it is an educated guess. It is hard to classify "what form does the differential equation have to be in" for the linear form to work. I get the trial solution only by assuming the left hand side to be zero for all $x$. It is also a byproduct of excluding the special case in the general solution in the second paragraph. It just happens that the constant term on the right hand side is $-1$ which matches the boundary condition $y(0)=1$. Otherwise, it does not work. I mention that at the beginning of the second paragraph. Please refer to the second paragraph for the general solution. – Hans May 21 '15 at 18:11