Please solve & show steps... # Solving Differential Equations with an Integrating Factor## Example ProblemShow that the differential equation $ x^2 y^3 + x(1 + y^2)y' = 0 $ is not exact, but becomes exact when multiplied by the integrating factor \[ \mu(x, y) = \frac{1}{x y^3} \]. Then solve the equation.### Exactness CheckThe given equation is not exact because\[ M_y = \underline{\hspace{2cm}} \]which is different from \[ N_x = \underline{\hspace{2cm}} \]### Multiplying by the Integrating FactorAfter multiplication with $ \mu(x, y) $, the equation is exact. The new equation satisfies:\[ M_y = N_x = \underline{\hspace{2cm}} \]### General SolutionThe general solution of the differential equation is given implicitly by\[ \underline{\hspace{4cm}} = c \]for any constant $ c $.### Explanation of Terms- $ M $: Function multiplied by $ dx $ in the original differential equation.- $ N $: Function multiplied by $ dy $ in the original differential equation.- $ M_y $: Partial derivative of $ M $ with respect to $ y $.- $ N_x $: Partial derivative of $ N $ with respect to $ x $.By ensuring that $ M_y = N_x $ after applying the integrating factor, we establish the exactness of the equation and proceed to solve for the general solution.

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Description: Please solve & show steps... # Solving Differential Equations with an Integrating Factor## Example ProblemShow that the differential equation $ x^2 y^3 + x(1 + y^2)y' = 0 $ is not exact, but becomes exact when multiplied by the integrating factor \

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xact, but becomes exact when multiplied by the integrating factor (x, y) 1 Then solve the equation. The given equation is not exact, because M,: vhich is different from N \fter multiplication with µ(x, y), the equation is exact, because then My = N = The general solution of the differential equation is given implicitly by = c, for any constant c.

xact, but becomes exact when multiplied by the integrating factor (x, y) 1 Then solve the equation. The given equation is not exact, because M,: vhich is different from N \fter multiplication with µ(x, y), the equation is exact, because then My = N = The general solution of the differential equation is given implicitly by = c, for any constant c.

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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# Solving Differential Equations with an Integrating Factor

## Example Problem

Show that the differential equation $ x^2 y^3 + x(1 + y^2)y' = 0 $ is not exact, but becomes exact when multiplied by the integrating factor
\[ \mu(x, y) = \frac{1}{x y^3} \]. Then solve the equation.

### Exactness Check
The given equation is not exact because
\[ M_y = \underline{\hspace{2cm}} \]
which is different from
\[ N_x = \underline{\hspace{2cm}} \]

### Multiplying by the Integrating Factor
After multiplication with $ \mu(x, y) $, the equation is exact. The new equation satisfies:
\[ M_y = N_x = \underline{\hspace{2cm}} \]

### General Solution
The general solution of the differential equation is given implicitly by
\[ \underline{\hspace{4cm}} = c \]
for any constant $ c $.

### Explanation of Terms
- $ M $: Function multiplied by $ dx $ in the original differential equation.
- $ N $: Function multiplied by $ dy $ in the original differential equation.
- $ M_y $: Partial derivative of $ M $ with respect to $ y $.
- $ N_x $: Partial derivative of $ N $ with respect to $ x $.

By ensuring that $ M_y = N_x $ after applying the integrating factor, we establish the exactness of the equation and proceed to solve for the general solution.

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