1st Order ODE Practice Questions
Problem 1:
Solve the following initial value problem:
y ′ = (2x + 3)(y 2 − 4), y(0) = −1
Solution:
The solution to the initial value problem is:
2
2 − 6e4x +12x
y=
1 + 3e4x2 +12x
Problem 2:
Solve the following initial value problem:
dy
x − y = x2 , y(1) = 2
dx
Solution:
The solution to the initial value problem is:
y = x2 + x
Problem 3:
Solve the following first order differential equation:
dy
− sin x · y = sin x
dx
Step 1: Rewrite in standard form
The equation is already in standard form:
dy
+ P (x)y = Q(x)
dx
where P (x) = − sin x and Q(x) = sin x.
Step 2: Find the integrating factor
The integrating factor µ(x) is given by:
R R
P (x) dx − sin x dx
µ(x) = e =e = ecos x
Step 3: Multiply through by the integrating factor
Multiply both sides of the equation by the integrating factor ecos x :
1
dy
ecos x
− ecos x sin x · y = ecos x sin x
dx
The left-hand side can be expressed as the derivative of yecos x :
d
(yecos x ) = ecos x sin x
dx
Step 4: Integrate both sides
Integrate both sides with respect to x:
Z
yecos x = ecos x sin x dx + C
Step 5: Solve for y
Now, solve for y:
Z
− cos x cos x
y=e e sin x dx + C
Step 6: Evaluate the Rintegral
To evaluate the integral ecos x sin x dx, we can use substitution. Let:
u = cos x ⇒ du = − sin x dx ⇒ −du = sin x dx
Thus, the integral becomes:
Z
eu (−du) = −eu + C = −ecos x + C
Final Answer
Substituting back into the equation for y, we have:
y = e− cos x (−ecos x + C)
Simplifying gives:
y = −1 + Ce− cos x
Thus, the final solution to the differential equation is:
y = Ce− cos x − 1
Problem 4: Solve the differential equation
2y ′ + y = 3x2 .
Solution:
1
y = 3x2 − 12x + 24 + Ce− 2 x .
Problem 5:
The differential equation is given by:
2
y ′ − y = 1 + 3 sin x,
with the initial condition:
y(0) = y0 .
Find the value of y0 for which the solution of the initial value problem
remains finite as t → ∞.
Answer:
−5
y0 =
2
Problem 6: Solve the first-order ordinary differential equation (ODE) given
by:
y ′ = e2x + y − 1.
Answer:
y = e2x + 1 + cex
Problem 6.1: The given differential equation in Problem 6 is not exact;
however, it can become an exact equation if multiplied by an appropriate in-
tegrating factor. Find this integrating factor and solve exact equation, then
compare results you obtained in Problem 6 and Problem 6.1.
Problem 7: Solve the following differential equation.
(2xy 2 + 2y) + (2x2 y + 2x)y ′ = 0.
Answer: The solution to the equation is
x2 y 2 + 2xy = c,
where c is a constant.
Problem 8: Solve the following differential equation:
p
xy ′ = 1 − y 2 .
Answer: The solution to the equation is
y = sin(ln |x| + c),
where x ̸= 0 and |y| < 1.
Problem 9: Solve the following differential equation:
y − 4x
y′ = .
x−y
Hint: This differential equation can be classified as separable if an appro-
priate change of variables is made.
3
Answer:
(y − 2x)(y + 2x)3 = exp(−4c)
Problem 10: Solve the following differential equation:
t2 y ′ + 2ty − y 3 = 0.
Hint: This differential equation is nonlinear; however, an appropriate change
of variables can transform it into a linear one.
Answer: The solution to the equation is
21
5t
y= ,
2 + 5ct5
where c is a constant.
Problem 11: Solve the following differential equation.
x
dx + ( − siny)dy = 0
y
Answer:
ν(y) = y; xy + ycosy − siny = c
