Solving a Separable Variable Differential Equation Using U-Substitution

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SUMMARY

The discussion focuses on solving the separable variable differential equation given by $$\dfrac{dy}{dx}=\dfrac{1}{7}\sqrt{y}\cos^2{\sqrt{y}}$$. Participants successfully separate variables and apply the substitution $$u=\sqrt{y}$$, leading to the integral $$2 \int \sec^2{u} \, du$$. This substitution simplifies the integration process, allowing for a clearer path to the solution. The use of u-substitution is confirmed as the next logical step in solving the differential equation.

PREREQUISITES
  • Understanding of separable differential equations
  • Familiarity with u-substitution in integration
  • Knowledge of trigonometric identities, specifically secant functions
  • Basic calculus skills, including integration techniques
NEXT STEPS
  • Study the method of u-substitution in greater detail
  • Practice solving separable differential equations with varying functions
  • Explore the properties and applications of the secant function in calculus
  • Learn about advanced integration techniques, including integration by parts
USEFUL FOR

Students and professionals in mathematics, particularly those studying differential equations, calculus, and integration techniques. This discussion is beneficial for anyone looking to enhance their problem-solving skills in mathematical analysis.

karush
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Solve the de $$\dfrac{dy}{dx}=\dfrac{1}{7}\sqrt{y}\cos^2{\sqrt{y}}$$
sepate variables
$$\displaystyle \dfrac{dy}{\sqrt{y}\, \cos^2{\sqrt{y}} }=\dfrac{1}{7}\, dx
\implies \int{\dfrac{{d}y}{\sqrt{y}\,\cos^2{\left(\sqrt{y} \right) }} }
= \int{ \dfrac{1}{7}\,{d}x}$$
ok i think u subst is next ... maybe...
$$u=\sqrt{y} \therefore du=\dfrac{{d}y}{2\,\sqrt{y}}$$
 
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karush said:
Solve the de $$\dfrac{dy}{dx}=\dfrac{1}{7}\sqrt{y}\cos^2{\sqrt{y}}$$
sepate variables
$$\displaystyle \dfrac{dy}{\sqrt{y}\, \cos^2{\sqrt{y}} }=\dfrac{1}{7}\, dx
\implies \int{\dfrac{{d}y}{\sqrt{y}\,\cos^2{\left(\sqrt{y} \right) }} }
= \int{ \dfrac{1}{7}\,{d}x}$$
ok i think u subst is next ... maybe...
$$u=\sqrt{y} \therefore du=\dfrac{{d}y}{2\,\sqrt{y}}$$
You stopped too soon! What did you get when you did the substitution?

-Dan
 
Last edited by a moderator:
using your sub \( u = \sqrt{y} \) ...

\( \displaystyle 2 \int \dfrac{1}{\cos^2{\sqrt{y}}} \cdot \dfrac{dy}{2\sqrt{y}} = 2\int \sec^2{u} \, du \)
 

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