Solve the given initial-value problem. Write the solution as...

Solve the given initial-value problem. Write the solution as an explicit algebraic function $y=f(x)$ (see Notes From the Classroom in Section 1.3). You may have to use a trigonometric identity.
$\left(1+x^{4}\right) \frac{d y}{d x}+x+4 x y^{2}=0, \quad y(1)=0$

Key Concepts

Initial Value Problems

An initial value problem is a differential equation paired with a specific condition that the solution must satisfy at a given point. This condition is used to determine the constants of integration after solving the differential equation, ensuring that the solution is unique and tailored to the scenario described.

Separable Differential Equations

A separable differential equation is one in which the variables can be divided so that all terms involving one variable are on one side of the equation and those involving the other variable are on the opposite side. This property simplifies the solution process by allowing each side to be integrated independently.

Integration Techniques

Integration techniques, including substitution and the use of standard inverse trigonometric integrals, are essential for solving the integrals obtained after separating the variables. Recognizing standard forms and making appropriate substitutions can greatly simplify the integration process and lead to an explicit solution.

Trigonometric Identities

Trigonometric identities, such as those used to simplify the expression involving the difference of angles, are vital tools in manipulating and rewriting expressions. These identities allow for the transformation of complex trigonometric expressions into simpler forms, facilitating the process of solving equations explicitly.

Transcript

00:01 We have to solve this initial value problem and express the final solution as an explicit function of y as a function of x.

00:08 So this is 1 plus x raised to 4, dy over dx, d .y over dx, and the other terms over to the right, so that's minus x minus 4x, y, square.

00:23 This can be read written as 1 plus x raised to 4, dy over dx, and if a minus x is taken out, we have 1 .6.

00:31 Plus 4 y square so let's do a cross multiplication here to bring the to separate the term so it's d y over 1 plus 4 y square is equal to minus x minus x d x over 1 plus x raised to 4 okay so we need to make some efforts here let's let's take this 1 over 4 let's take this 4 out so we are left that 1 over 4 and in here we are left with d .y over y square plus 1 over 4.

01:06 This is equal to, this can be written as minus x, t, x, over 1 plus x raise to 4 can be written as x square, whole square.

01:17 Now we are going to make a substitution of x square as t.

01:23 This means that 2x dx is dt, which also means that x, dx is dt over 2.

01:35 So if you make that substitution, let's continue the green one again.

01:39 So 1 over 4 remains as it is.

01:41 This is dy over y square plus 1 over 4 can be written as 1 over 2 whole square.

01:47 This is equal to x, t, x, x is replaced by dt over 2 over 2 over this is 1 plus t square.

01:53 And now this can be integrated easily.

01:56 Integration of 1 over y square plus 1 over 2 whole square is 1 over 2, tan, inverse, y over 2.

02:05 Is equal to minus 1 over 2 is a constant...

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