# Solve the initial value problem y' = x 3 ( 1 - y ) , y ( 0 ) = 3

## Question:

Solve the initial value problem {eq}y'=x^3(1-y), \ \ \ y(0) =3 {/eq}

## Separable Differential Equation

This differential equation is the type where the variables can be separated and so the equation can be written in the form

{eq}P(y)\;dy = Q(x)\;dx {/eq}

This is solved by taking the integral of both sides, that is

{eq}\displaystyle \int P(y)\;dy = \int Q(x)\;dx {/eq}

## Answer and Explanation:

The differential equation given has variables that can be separated.

{eq}y'=x^3(1-y) \\ \dfrac{dy}{dx} = x^3(1-y) \\ \dfrac{1}{(1-y)}\;dy = x^3\;dx {/eq}

This type of equation is called a first order separable differential equation and can be solved by integrating both sides of the equation.

{eq}\int \dfrac{1}{(1-y)}\;dy = \int x^3\;dx {/eq}

For the left side of the equation, we can apply substitution method to solve the integral. Let

{eq}u = 1-y \\ du = -dy \rightarrow dy = -du {/eq}

The integral becomes

{eq}\int \dfrac{1}{u}\;(-du) = \int x^3\;dx \\ \int -\dfrac{1}{u}\;du = \int x^3\;dx \\ -\int \dfrac{1}{u}\;du = \int x^3\;dx \\ -\ln(u) = \int x^3\;dx {/eq}

Substitute back the value of u.

{eq}-\ln(1-y) = \int x^3\;dx {/eq}

For the right side of the equation, we can apply the power rule {eq}\int x^n \;dx = \frac{x^{n+1}}{n+1} {/eq}.

{eq}-\ln(1-y) = \dfrac{x^{3+1}}{3+1} \\ -\ln(1-y) = \dfrac{x^{4}}{4} \\ \ln(1-y) = -\dfrac{x^{4}}{4} {/eq}

Add a constant of integration to the equation.

{eq}\ln(1-y) = -\dfrac{x^{4}}{4} + C {/eq}

Write the equation as {eq}y {/eq} in terms of {eq}x {/eq}.

{eq}e^{\ln(1-y)} = e^{-\frac{x^{4}}{4} + C} \\ (1-y) = e^{-\frac{x^{4}}{4} + C} \\ (y-1) = -e^{-\frac{x^{4}}{4} + C} \\ y = -e^{-\frac{x^{4}}{4} + C} + 1 \\ y(x) = -e^{-\frac{x^{4}}{4}}\cdot e^C + 1 {/eq}

Since {eq}C {/eq} is just an arbitrary constant, we know that {eq}e^C {/eq} will also be a constant and we can write it as {eq}K {/eq}.

{eq}y(x) = -Ke^{-\frac{x^{4}}{4}} + 1 {/eq}

We can solve for the value of the unknown constant using the condition given.

{eq}y(0) = -Ke^{-\frac{(0)^{4}}{4}} + 1 = 3 \\ -Ke^{-\frac{0}{4}} = 3 - 1 \\ -Ke^{0} = 2 \\ -K(1) = 2 \\ -K = 2 \\ K = -2 \\ y(x) = -(-2)e^{-\frac{x^{4}}{4}} + 1 {/eq}

The complete solution to the differential equation is

{eq}\boldsymbol{y(x) = 2e^{-\frac{x^{4}}{4}} + 1} {/eq}

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from GRE Math: Study Guide & Test Prep

Chapter 16 / Lesson 1