Question

If y₁ = 2x+2 and y₂= -x²/2 are the solutions of the equation y= xy´+( y´)²/2 ,
Then are the constant multiples c₁y₁ and c₂y₂ , where c₁,c₂ are arbitary, also the solutions of the given DE ? Is the sum y₁+y₂ a solution ? Justify your answer .

Answer 1

1)Let's check, if $c_1y_1$ is a solution:

$c_1y_1=2c_1(x+1).$ .

Let's check, if $2c_1(x+1)$ is equal to $x*(2c_1(x+1))'+\cfrac{((2c_1(x+1))')^2}{2}$ :

$x*(2c_1(x+1))'+\cfrac{((2c_1(x+1))')^2}{2} = 2c_1x+2c_1^2$ and it's not equal to $2c_1(x+1)$ for arbitary $c_1$ so it's not a solution.

2)Let's check if $c_2y_2$ is a solution:

$c_2y_2=\cfrac{-c_2x^2}{2}$ .

Let's check, if $\cfrac{-c_2x^2}{2}$ is equal to $x*(\cfrac{-c_2x^2}{2})'+\cfrac{((\cfrac{-c_2x^2}{2})')^2}{2}:$

$x*(\cfrac{-c_2x^2}{2})'+\cfrac{((\cfrac{-c_2x^2}{2})')^2}{2}=-c_2x^2+\cfrac{c_2^2x^2}{2}$ and it's not equal to $\cfrac{-c_2x^2}{2}$ for arbitary $c_2$ so it's not a solution.

3)Let's check, if $y_1+y_2$ is a solution:

$(y_1+y_2)'=(\cfrac{-x^2}{2}+2x+2)'=2-x$ .

So $xy´+\cfrac{( y´)²}{2}=2x-x^2+\cfrac{1}{2}$ and it's not equal to $y_1+y_2$ .

Answer: none of the three expressions is a solution.