Question

If y1 = 2x + 2and y2 = −x^2/2 are the solutions of the equation y = x y′+(y')^2/ 2then are the constant multiples c1y1and c2y2 , where c1, c2 are arbitrary, also the solutions of the given de? is the sum y1+ y2 a solution? Justify your answer.

Answer 1

Let

$$y_1=2x+2, y_2=-{x^2 \over 2}$$

Consider

$$y^*=c_1y_1+c_2y_2$$

$$y^*=c_1(2x+2)+c_2(-{x^2 \over 2})=(-{c_2 \over 2})x^2+(2c_1)x+2c_1$$

where c1, c2 are arbitrary constants.

Then

$$(y^*)'=c_1(y_1)'+c_2(y_2)'=c_1(2x+2)'+c_2(-{x^2 \over 2})'=2c_1-xc_2$$

Consider

$$x(y^*)'+{((y^*)')^2 \over 2}$$

$$x(2c_1-xc_2)+{(2c_1-xc_2)^2 \over 2}=2xc_1-x^2c_2+2c_1^2-2xc_1c_2+{1 \over 2}x^2c_2^2=$$

$$=({1 \over 2}c_2^2-c_2)x^2+(2c_1-2c_1c_2)x+2c_1^2$$

The necessary conditions are

$${1 \over 2}c_2^2-c_2=-{c_2^2 \over 2}$$ $$2c_1-2c_1c_2=2c_1$$ $$2c_1^2=2c_1$$

$$c_2(c_2-1)=0$$ $$c_1c_2=0$$ $$c_1(c_1-1)=0$$

If $c_1=0,$ then $c_2=0$ or $c_2=1$.

If $c_2=0,$ then $c_1=0$ or $c_1=1.$

We have three pairs for $(c_1, c_2)$

$$(0, 0), (0, 1), (1, 0)$$

The sum $y_1+y_2$ is not the solution of the differential equation.