Separable differential equations and initial conditions — 9709 P3
Separate a first-order equation and use an initial condition and branch.
Notes and key methods
A separable equation can be written h(y)dy=g(x)dx. Integrate both sides and include one arbitrary constant. If division by a function of y is used, inspect any excluded constant solutions separately.
Use the initial condition to fix the constant and any branch. A square-root solution must keep the sign required by the initial value and remain in a domain where the original differential equation is defined.
Check the conditions
Separate variables before integrating.
Use the initial condition to select the valid branch.
Three worked examples
Example 1 · Separating variables
For y>0, \(\frac{dy}{dx}=\frac{4x}{y}\), with \(y(0)=5\). For the short checks k=4, a=5.
Solve the initial-value problem explicitly, stating the branch.
Show that the explicit solution satisfies the original differential equation.
Find the positive x where y=10.
Find C in the relation y²=kx²+C.
Show the reasoning
(a)
Separate as y dy=kx dx. Integration gives y²/2=kx²/2+C₀. The initial condition gives y²=kx²+a². Since y(0)>0, use the positive square root.
Answer: \(y=\sqrt{4 x^{2} + 25}\)
(b)
Differentiating the positive square root gives kx/√(kx²+a²)=kx/y. Its value at zero is a and its denominator is positive for every real x.
Answer: \(y'=kx/y\) with \(y(0)=5\).
(c)
Substitute the stated y into the squared relation, solve for x² and keep the positive x specified.
Answer: \(\frac{5 \sqrt{3}}{2}\)
(d)
Apply y(0)=a.
\(25\)
Answer: \(25\).
Example 2 · Separating variables
For y>0, \(\frac{dy}{dx}=\frac{1x}{y}\), with \(y(0)=6\). For the short checks k=1, a=6.
Solve the initial-value problem explicitly, stating the branch.
Show that the explicit solution satisfies the original differential equation.
Find the positive x where y=12.
Find C in the relation y²=kx²+C.
Show the reasoning
(a)
Separate as y dy=kx dx. Integration gives y²/2=kx²/2+C₀. The initial condition gives y²=kx²+a². Since y(0)>0, use the positive square root.
Answer: \(y=\sqrt{x^{2} + 36}\)
(b)
Differentiating the positive square root gives kx/√(kx²+a²)=kx/y. Its value at zero is a and its denominator is positive for every real x.
Answer: \(y'=kx/y\) with \(y(0)=6\).
(c)
Substitute the stated y into the squared relation, solve for x² and keep the positive x specified.
Answer: \(6 \sqrt{3}\)
(d)
Apply y(0)=a.
\(36\)
Answer: \(36\).
Example 3 · Separating variables
For y>0, \(\frac{dy}{dx}=\frac{2x}{y}\), with \(y(0)=2\). For the short checks k=2, a=2.
Solve the initial-value problem explicitly, stating the branch.
Show that the explicit solution satisfies the original differential equation.
Find the positive x where y=4.
Find C in the relation y²=kx²+C.
Show the reasoning
(a)
Separate as y dy=kx dx. Integration gives y²/2=kx²/2+C₀. The initial condition gives y²=kx²+a². Since y(0)>0, use the positive square root.
Answer: \(y=\sqrt{2 x^{2} + 4}\)
(b)
Differentiating the positive square root gives kx/√(kx²+a²)=kx/y. Its value at zero is a and its denominator is positive for every real x.
Answer: \(y'=kx/y\) with \(y(0)=2\).
(c)
Substitute the stated y into the squared relation, solve for x² and keep the positive x specified.
Answer: \(\sqrt{6}\)
(d)
Apply y(0)=a.
\(4\)
Answer: \(4\).
Common mistakes
Do not lose a factor of two when integrating y dy.
An implicit squared relation can contain an invalid branch.