Find the particular solution ofthe differential equation y3(x4 + 1)y′ − x3( y4 + 1) = 0 that satisfiesthe initial condition y(0) = 1.

Answered: Find the particular solution of the…

Calculus For The Life Sciences

2nd Edition

ISBN:9780321964038

Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.

Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.

Chapter11: Differential Equations

Section11.CR: Chapter 11 Review

Problem 33CR

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Question

Find the particular solution of

the differential equation y³(x⁴ + 1)y′ − x³( y⁴ + 1) = 0 that satisfies

the initial condition y(0) = 1.

Expert Solution

Step 1

the given differential equation is:

$y^{3} (x^{4} + 1) y' - x^{3} (y^{4} + 1) = 0$

we have to find the particular solution of the given differential equation that satisfies the initial condition y(0)=1.

the differential equation can be written as:

$\begin{array}{rcl} y^{3} (x^{4} + 1) y' - x^{3} (y^{4} + 1) & = & 0 \\ y^{3} (x^{4} + 1) \frac{d y}{d x} & = & x^{3} (y^{4} + 1) \\ \frac{y^{3}}{y^{4} + 1} \frac{d y}{d x} & = & \frac{x^{3}}{x^{4} + 1} \\ \frac{y^{3}}{y^{4} + 1} d y & = & \frac{x^{3}}{x^{4} + 1} d x \end{array}$

Step 2

now integrate the both the sides of the equation,

therefore

$\int \frac{y^{3} d y}{y^{4} + 1} = \int \frac{x^{3} d x}{x^{4} + 1} (1)$

let the integral $\int \frac{y^{3} d y}{y^{4} + 1}$ be I.

therefore,

$I = \int \frac{y^{3} d y}{y^{4} + 1}$

let $y^{4} + 1 = t$

therefore,

$\begin{array}{rcl} d (y^{4} + 1) & = & d t \\ 4 y^{3} d y & = & d t \\ y^{3} d y & = & \frac{d t}{4} \end{array}$

Step 3

substitute these values in the integral I.

therefore,

$\begin{array}{rcl} I & = & \int \frac{y^{3} d y}{y^{4} + 1} \\ = & \int \frac{d t}{4 t} \\ = & \frac{1}{4} \int \frac{d t}{t} \\ = & \frac{1}{4} \ln (t) \end{array}$

now substitute the value of t that is $t = y^{4} + 1$ in the integral I.

therefore,

$\begin{array}{rcl} I & = & \frac{1}{4} \ln (t) \\ = & \frac{1}{4} \ln (y^{4} + 1) \end{array}$

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Find the particular solution of the differential equation y3(x4 + 1)y′ − x3( y4 + 1) = 0 that satisfies the initial condition y(0) = 1.