Construction of solutions to Cauchy problem from existing solutions.

Question

Please help me to handle 3 following problems:

1. Equation $y' = 2y+f(x)$ has the solution $g(x)$ such that $g(0)=1$. Find the solution $u(x)$ to $y(0)=0$ using $g(x)$.
2. Homogenous equation $y'=f(x,y)$ has the solution $g(x)$ such that $g(1)=2$. Find the solution $u(x)$ to $y(2)=4$ using $g(x)$.
3. Equation $y' = a(x)y+b(x)y^2$ has the solutions $g(x)$ and $f(x)$ such that $g(0)=1$ and $f(0)=2$. Find the solution $u(x)$ to $y(0)=3$ using $g(x)$ and $f(x)$.

My attempts:

1. $u(x)=g(x)-1$ is incorrect answer of course. Correct answer is $u(x)=g(x)-e^{2x}$. I solved it by looking for $u(x)$ in form $u(x) = g(x) + h(x)$ and used initial conditions to obtain $h(x)$.
2. $u(x) = 2\,f(x)=f(x)+f(x)$. Since the equation is homogenous $u(x)$ is solution and it satisfied initial conditions.
3. Tempting answer $u(x) = g(x)+f(x)$ looks incorrect to me, since the equation is non-linear and method I used in problem 1 fails.

My questions:

1. Are my solutions for problem $1$ and $2$ valid? I am sure it is, but I have to ask it since if I made a mistake it can help me with problem $3$.
2. Please advice how to handle problem $3$. I am sorry I have no idea how to approach it. Is it true that $u(x) = g(x)+f(x)$ is not the solution?

Thanks a lot for your help!

Answer 1

1. You can use the fact the the solution to the Cauchy problem $$ y' = 2y + f(x), \qquad y(0) = x_0, $$ is given by $$ y(x) = e^{2x}\left[x_0 +\int_0^x f(s)\, ds\right]. $$ For $x_0=1$ you have $g(x)$, and for $x_0=0$ you have $u(x)$.

Also linearity can be directly invoked without using this representation formula for solutions, since you know that the general integral of the equation is of the form $y(x) = v(x) + g(x)$, where $v$ is a solution of the homogeneous equation $y' = 2y$ (and $g$ is any solution of the non-homogeneous equation).

2. Since $f(\lambda x, \lambda y) = f(x,y)$, for every $\lambda\neq 0$ the function $u(x) = g(\lambda x) / \lambda$ is a solution of the same equation: $$ u'(x) = g'(\lambda x) = f(\lambda x, g(\lambda x)) = f(\lambda x, \lambda u(x)) = f(x, u(x)). $$ You can now choose $\lambda = 1/2$.

3. Since $y(x) = 0$ is a solution of the equation, by uniqueness you have that $f$, $g$ and $u$ are strictly positive. Hence, using the transformation $z=1/y$, the functions $1/f$, $1/g$, $1/u$ are solutions of the linear differential equation $$ (*)\qquad z' = -a(x) z - b(x). $$ Since $1/f - 1/g$ is a solution of the homogeneous equation $z'=-a(x) z$, and $1/g$ is a solution of (), the general solution of () is of the form $$ z = (1/f - 1/g) c + 1/g. $$ Since you are looking for the solution satisfying $z(0) = 1/u(0) = 1/3$, you can easily compute the value of $c = 4/3$.