Solve the initial value problem $y'+y=y^2, y(0)=-1$

Question

I want to solve this problem by using Laplace transform. But I don't know how to transform ${\cal L}[y^2]$. Does anyone has any idea by using Laplace or else? Thanks.

Answer 1

$$y'+y=y^2 \implies y^{-2}y'+y^{-1}=1$$ Let $y^{-1}=v \implies y^{-2} y'=-v'$, thenwe get $$v'=v-1 \implies \int \frac{dv}{v-1}=\int dx+C$$ You may complete it now as $$\frac{1}{y(x)}=De^x+1 \implies y(x)=\frac{1}{De^{x}+1}\implies D=-2$$ Finally $$y(x)=\frac{1}{1-2e^x}$$

Answer 2

The OP says "solve using the Laplace transform" ...

As pointed out by Hans, the Laplace transform is no good for nonlinear DEs. We can rearrange this DE, though. Instead of solving for $y$ as a function of $x$, try to solve for $x$ as a function of $y$. $$ \frac{dy}{dx}+y=y^2 \\ \frac{1}{dx/dy} + y = y^2 \\ \frac{dx}{dy} = \frac{1}{y^2-y} $$ Now it is a linear DE, so susceptible for Laplace transform methods.

Next problem: because of the poles at $y=0$ and $y=1$, the Laplace transform of this right-hand side does not exist (as a function). You could try interpreting that Laplace transform as a Schwartz distribution; I won't do that here.

Let's translate the variable: $z=y-2$ so that the poles are at $z=-1, z=-2$ and the Laplace transform will exist. $$ \frac{dx}{dz} = \frac{1}{z^2+3z+2} = \frac{1}{z+1}-\frac{1}{z+2} $$ Now the Laplace transform $X(t) = \mathscr{L}(x(z))$ satisfies (using Laplace transform tables) $$ tX(t)-x(0) = e^t\operatorname{Ei}_1(t)-e^{2t}\operatorname{Ei}_1(2t) $$ where $x(0)$ is $x$ when $z=0$, that is, when $y=2$. Let's call this value $c$. Also note: $\operatorname{Ei}_1$ is an exponential integral function; something we find in our Laplace transform tables. Solve: $$ X(t) = \frac{e^t\operatorname{Ei}_1(t)-e^{2t}\operatorname{Ei}_1(2t)+c}{t} $$ Take inverse Laplace transform (again using tables): $$ x(z) = c-\ln\frac{z+2}{2(z+1)} $$ Although this is guaranteed only for $z>0$ or $y>2$, let's proceed anyway: with $z=y-2$, we get $$ x = c - \ln\frac{y}{2(y-1)} $$ The given initial condition says when $y=-1$ we have $x=0$. Solve to get $c=-2\ln 2$. Then $$ x = -2\ln 2 - \ln\frac{y}{2(y-1)} = \ln \frac{y-1}{2y} \\ e^x = \frac{y-1}{2y} \\ y=\frac{1}{1-2e^x} $$ Plug into the original DE to verify that this is, indeed, the solution.

Answer 3

I would do this a little differently. From $\frac{dy}{dx}+ y= y^2$ we have $\frac{dy}{dx}= y^2- y$. Separating variables, $\frac{dy}{y^2- y}= dx$

$y^2- y= y(y- 1)$ so we can write the right side as "partial fractions" $\frac{1}{y(y-1}= \frac{A}{y}+ \frac{B}{y- 1}$ Multiply on both sides by y(y- 1) $1= 0y+ 1= A(y- 1)+ By= (A+ B)y- A$.

So we have 0= A- B and 1= -A. A= -1 so -1- B= 0 and B= -1.

$\frac{dy}{y(y- 1)}= -\frac{dy}{y}- \frac{dy}{y- 1}$

Integrating -ln(y)- ln(y- 1)= -ln(y(y- 1))= -ln(y^2- y)= x+ C.

Taking the exponential of both sides, $\frac{1}{y^2- y}= C'e^x$ where $C'= e^C$.

Answer 4

$\frac{dy}{dx}=y^2-y \implies \int \frac{dy}{y^2-y}=\int dx$. We get $\ln \frac{y-1}{y}=x+C$. Upon simplification, we get $y=\frac{1}{1-2e^x}$.