2. Solve initial value problem $\frac{dy}{dx} = \frac{y \cos x}{1+2y^2}$, $y(0) = 1$
Added by Carolina C.
Close
Step 1
We can rewrite the equation as $\frac{1+2y^2}{y} dy = \cos x dx$. Show more…
Show all steps
Your feedback will help us improve your experience
Nick Johnson and 95 other Calculus 1 / AB educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Solve the given initial-value problem. $$\left(\frac{1}{1+y^{2}}+\cos x-2 x y\right) \frac{d y}{d x}=y(y+\sin x), \quad y(0)=1$$
First-Order Differential Equations
Exact Equations
Solve the given initial-value problem. $$ \left(\frac{1}{1+y^{2}}+\cos x-2 x y\right) \frac{d y}{d x}=y(y+\sin x), \quad y(0)=1 $$
Find the solution of the given initial value problem. $$ \frac{d y}{d x}=2 x \cos (y(x)) \quad y(0)=0 $$
Applications of the Integral
First Order Differential Equations—Separable Equations
Recommended Textbooks
Calculus: Early Transcendentals
Thomas Calculus
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD