Chapter 6: Q. 48 (page 570)

Use antidifferentiation and/or separation of variables to solve each of the initial-value problems in Exercises 29-52.
$\frac{d y}{d x} = x y + x + y + 1, y (0) = c$

Short Answer

Expert verified

On solving, we get $y (x) = - 1 + (1 + c) e^{\frac{1}{2} (x + 1)^{2}}$

Step by step solution

Step 1. Given information

Given the expression $\frac{d y}{d x} = x y + x + y + 1, y (0) = c$

Take common and use variable separable method

Calculating, we get

$\frac{d y}{d x} = x (y + 1) + 1 (y + 1) = (x + 1) (y + 1)$

Integrating, we get

$\int \frac{1}{y + 1} d y = \int (x + 1) d x \ln | y + 1 | = \frac{1}{2} (x + 1)^{2} + C y + 1 = e^{\frac{1}{2} (x + 1)^{2} + C} y = - 1 + A e^{\frac{1}{2} (x + 1)^{2}}$

Substitute x=0,y=c in the equation and solve

Calculating, we get

$c = - 1 + A A = 1 + c y (x) = - 1 + (1 + c) e^{\frac{1}{2} (x + 1)^{2}}$

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Use antidifferentiation and/or separation of variables to solve each of the initial-value problems in Exercises 29-52.
$\frac{d y}{d x} = x y + x + y + 1, y (0) = c$

Short Answer

Step by step solution

Step 1. Given information

Take common and use variable separable method

Substitute x=0,y=c in the equation and solve

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Use antidifferentiation and/or separation of variables to solve each of the initial-value problems in Exercises 29-52.dydx=xy+x+y+1,y(0)=c

Short Answer

Step by step solution

Step 1. Given information

Take common and use variable separable method

Substitute x=0,y=c in the equation and solve

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Probability and Statistics

Geometry

Mechanics Maths

Decision Maths

Applied Mathematics

Calculus

Study anywhere. Anytime. Across all devices.

Use antidifferentiation and/or separation of variables to solve each of the initial-value problems in Exercises 29-52.
$\frac{d y}{d x} = x y + x + y + 1, y (0) = c$