Find the particular solution that satisfies the given initial condition(y2+y)exy′=y3+exy3;y(0)=1\left(y^2+y\right)e^{x}y^{\prime}=y^3+e^{x}y^3;y\left(0\right)=1(y2+y)exy′=y3+exy3;y(0)=1

ln ⁡ ∣ y ∣ − 1 y = − e − x + x \ln\left|y\right|-\frac{1}{y}=-e^{-x}+x ln ∣ y ∣ − y 1 = − e − x + x

Find the particular solution that satisfies the given initial condition ( y 2 + y ) e x y ′ = y 3 + e x y 3 ; y ( 0 ) = 1 \left(y^2+y\right)e^{x}y^{\prime}=y^3+e^{x}y^3;y\left(0\right)=1 ( y 2 + y ) e x y ′ = y 3 + e x y 3 ; y ( 0 ) = 1

we're asked to find the particular solution that satisfies the given initial condition, and we're doing this based on the differential equation y squared plus y times e to the x y prime equals y cubed plus e to the x times y cubed. The initial condition that we're given is that y of zero is equal to one. So given this differential equation, this looks rather complicated, but it is a separable differential equation. I'm going to go ahead and rewrite it down here, replacing that y prime with d y d x. So I have y squared plus y times e to the x times d y d x, and that's equal to y cubed plus e to the x y cubed. So now what we want to do here is get all of our functions of y, which show up in multiple places on the same side with d y, and then everything involving x and d x together on the other side. Now this might look like it's going to be rather complicated here, but I can already see something that's going to help me out. Both of these terms have a y cubed, so I can easily factor that. So I can rewrite this as y squared plus y times e to the x, d y d x equals y cubed times one plus e to the x. So now this makes it easier for me to continue separating my variables because I have most of my y terms together on this side, and I can easily move this term over by dividing both sides by y cubed in order to cancel that out. So now I have y squared plus y over y cubed times e to the x dy dx equals one plus e to the x. So now I want to move my x terms over to the side the other side, so that's e to the x and that dx on the bottom there. So I can divide both sides by e to the x in order to get that to cancel, and then at the same time I can multiply both sides by dx as well. So now I have y squared plus y over y cubed times dy equals one plus e to the x over e to the x d x. So I've already successfully separated my variables, but there are still some things I can do here to make this a bit easier. Because we only have one term in our denominators, we can split this out into multiple fractions, which I know will end up helping me later because we're gonna have to integrate. So I'm gonna go ahead and split this up into y squared over y cubed plus y over y cubed, and all of that is being multiplied by dy. Then on this side I have one over e to the x plus e to the x over e to the x times dx. So we can see some stuff is going to cancel here. Now over here on the left side where I have all my y terms, this y squared will cancel out two of those y cubes on the bottom there, and then this one y will bring this exponent down to a two. So this leaves me with one over y plus one over y squared dy, and that's equal to one over e to the x. And then these e to the x's will just cancel each other out, so I have plus one times a d x. So now let's go ahead and proceed on with step two here, which is integrating both sides. Now that I've already split out my fractions, I've made this easier for me to integrate. So integrating both sides here. Remember, we're integrating each term individually, so one over y, since they're being added together, I can just focus on each of these terms. So one over y, that's going to give me the natural log of the absolute value of y. And then that y squared, I can apply my power rule here, and this is going to end up giving me one of negative one over y. So I can write that as minus one over y. Then this is equal to one over e to the x, so the integral of this term is going to be a negative e to the negative x, and then plus x, and then I'll add on my constant of integration over here, c. So I've finished that second step here, and we were asked to find a particular solution based on this initial condition that y of zero is equal to one. So that means I need to go ahead and proceed on to step three, which is finding c by plugging that initial condition in where x is zero and y is one. So let's come back down here and do that. So if x is zero and y is one, I'm gonna write this as the natural log of one minus one over one equals negative e to the negative zero plus zero plus c. So I've just plugged in that initial condition, and my goal here is to solve for that constant of integration c. Now e to the power of zero is going to be one. This is negative, so this ends up being negative one. And then natural log of one is zero, so that will go away. One over one is one, so on this side I have negative one and that's equal to negative one plus c. Now if I add one on both sides here, that will end up canceling out and it actually ends up canceling out on both sides. So I'm left with c actually being equal to zero. So that means that whatever we're solving for y and getting our final particular solution, our constant is just zero. So with that in mind, coming back up to this original equation here, we're plugging that constant in and and I'm going to go ahead and do that to the side. So this is the natural log of the absolute value of y minus one over y, and that's equal to negative e to the negative x plus x and then plus zero so I don't have to worry about writing that. Now typically in this final step, we would be solving for y, but looking at this equation, there's no way that I can get y all by itself because it shows up in both a natural log and in the denominator of a fraction in a totally different term. And so we can't actually solve for y here because this is an implicit equation. We can't get that y by itself, so this is actually our final solution here, that the natural log of the absolute value of y minus one over y is equal to negative e to the negative x plus x, and we're all done here.

Find the particular solution that satisfies the given initial condition ( y 2 + y ) e x y ′ = y 3 + e x y 3 ; y ( 0 ) = 1 \left(y^2+y\right)e^{x}y^{\prime}=y^3+e^{x}y^3;y\left(0\right)=1 ( y 2 + y ) e x y ′ = y 3 + e x y 3 ; y ( 0 ) = 1

ln ⁡ ∣ y ∣ − 1 y = − e − x + x \ln\left|y\right|-\frac{1}{y}=-e^{-x}+x ln ∣ y ∣ − y 1 = − e − x + x

ln ⁡ ∣ y ∣ − 1 y = − e − x + x + 1 \ln\left|y\right|-\frac{1}{y}=-e^{-x}+x+1 ln ∣ y ∣ − y 1 = − e − x + x + 1

ln ⁡ ∣ y y 2 ∣ = − e − x + x \ln\left|\frac{y}{y^2}\right|=-e^{-x}+x ln y 2 y = − e − x + x

ln ⁡ ∣ y y 2 ∣ = − e − x + x + 1 \ln\left|\frac{y}{y^2}\right|=-e^{-x}+x+1 ln y 2 y = − e − x + x + 1

13: Intro to Differential Equations

Separable Differential Equations

Business Calculus

13: Intro to Differential Equations

Separable Differential Equations

Struggling with Business Calculus?

Join thousands of students who trust us to help them ace their exams!Watch the first video

Multiple Choice

Find the particular solution that satisfies the given initial condition
$\left(y^2+y\right)e^{x}y^{\prime}=y^3+e^{x}y^3;y\left(0\right)=1$

$\ln\left|y\right|-\frac{1}{y}=-e^{-x}+x$

$\ln\left|y\right|-\frac{1}{y}=-e^{-x}+x+1$

$\ln\left|\frac{y}{y^2}\right|=-e^{-x}+x$

$\ln\left|\frac{y}{y^2}\right|=-e^{-x}+x+1$

Verified step by step guidance

Step 1: Start by rewriting the given differential equation $(y^2 + y)e^x y' = y^3 + e^x y^3$ in a more manageable form. Factor out common terms where possible. Notice that $y^3$ is a common term on the right-hand side.

Step 2: Divide through by $(y^2 + y)e^x$ to isolate $y'$ on one side. This will give you $y' = \frac{y^3 + e^x y^3}{(y^2 + y)e^x}$. Simplify the expression by factoring and canceling terms where appropriate.

Step 3: Recognize that this is a separable differential equation. Rearrange the terms to separate $y$ and $x$, so that all $y$-dependent terms are on one side and all $x$-dependent terms are on the other. This will involve dividing and simplifying the terms.

Step 4: Integrate both sides. For the $y$-dependent side, you will need to integrate an expression involving $\frac{1}{y}$ and $\frac{1}{y^2}$. For the $x$-dependent side, you will integrate $e^{-x}$ and any constants. Use the initial condition $y(0) = 1$ to solve for the constant of integration.

Step 5: After integrating, simplify the resulting expression to match the form of the solution. Combine logarithmic terms and constants as needed. The final solution will involve $\ln|y|$, $\frac{1}{y}$, and terms involving $e^{-x}$ and $x$.

7:15m

Watch next

Master Separation of Variables with a bite sized video explanation from Patrick

Start learning

Separable Differential Equations practice set

Problem sets built by lead tutors

Expert video explanations

Go to Practice