Question

Solve the following bvp by eigenfunction expansion method.

y^''+2y=2-4x, y(0)=0, y(1)+y^'(1)=0

Answer 1

Characteristics equation:

$k^2+2=0\\roots\space are\\ k_{1,2}=\pm\sqrt{2}\\$

Basic solutions:

$y_2(x)=cos(\sqrt{2}\cdot x)\\ y_1(x)=sin(\sqrt{2}\cdot x)\\$

General solution of homogeneous part:

$y_h(x,C_1,C_2)=C_1\cdot cos(\sqrt{2}\cdot x)+ C_2\cdot sin(\sqrt{2}\cdot x);$

Let find partial solution in form:

$y_1=a+b\cdot x$

For finding a,b substitute it in diff equation:

$y_1''+2\cdot y_1=2a+2bx=2-4x$

We take a,b values:

a=1, b=-2;

Partial solution has the form:

$y_1(x)=1-2\cdot x$

The overall solution as a whole

$y_{nh}(x,C_1,C_2)=C_1\cdot cos(\sqrt{2}\cdot x)+ C_2\cdot sin(\sqrt{2}\cdot x)+1-2\cdot x$

Now we apply boundary condition y(0)=0 and have:

0=$C_1+1$

From it we have $C_1=-1$

And therefore

$y(x,C_2)= -cos(\sqrt{2}\cdot x)+ C_2\cdot sin(\sqrt{2}\cdot x)+1-2\cdot x$

Further we use second boundary condition:

$y(1)+y'(1)=-cos(\sqrt 2)+C_2\cdot sin(\sqrt 2)-1+\sqrt 2\cdot \\ sin(\sqrt 2)+\sqrt 2\cdot C_2\cdot cos(\sqrt 2)-2=0$

This is an equation with respect to C₂

Solving it we have:

$C_2=\frac{3+cos(\sqrt 2)-\sqrt 2\cdot sin(\sqrt 2)}{sin(\sqrt 2)+\sqrt 2\cdot cos(\sqrt 2)}$

And finally

$y(x,)= -cos(\sqrt{2}\cdot x)+\\ \left( \frac{3+cos(\sqrt 2)+\sqrt 2\cdot sin(\sqrt 2)}{sin(\sqrt 2)+\sqrt 2\cdot cos(\sqrt 2)} \right )\cdot sin(\sqrt{2}\cdot x)+1-2\cdot x$

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