Find the particular solution to the differential equation y′=2sinx+3cosxy^{\prime}=2\sin x+3\cos x given the initial condition y(0)=4y\left(0\right)=4.

y = − 2 cos ⁡ x + 3 sin ⁡ x + 6 y=-2\cos x+3\sin x+6

Find the particular solution to the differential equation y ′ = 2 sin x + 3 cos x y^{\prime}=2\sin x+3\cos x given the initial condition y ( 0 ) = 4 y\left(0\right)=4 .

Here we're asked to find the particular solution to the differential equation y prime equals two sine of x plus three cosine of x given the initial condition that y of zero is equal to four. So given that y prime is equal to two sine x plus three cosine x, that means that we can go ahead and integrate here to get that y is equal to the integral of two sine x plus three cosine x d x. X. So now we just want to find this integral. So integrating that first term to sine of x, this integral will give me negative two cosine of x, and then plus three sine of x for that next term. Then we wanna go ahead and add on our constant of integration c there since this is just our general solution based on the indefinite integral. So this gives us our general solution that y is equal to negative two cosine of x plus three sine of x plus c. But we need to take this a step further because we're asked to find the particular solution given that y of zero is equal to four. So with this particular solution, we wanna go ahead and plug in zero for x and four for y into our general solution so that we can find this constant and write our particular solution. So if four is equal to negative two cosine of zero, plus three sine of zero, plus C. Let's solve for C here. Now the cosine of zero here, thinking about our unit circle, this is going to be equal to one. So this gives me four equals negative two times one and then plus three sine of zero, sine of zero is zero. So this whole term goes away. It's just zero and I'm just left with that plus c. Now negative two times one is just negative two, so I can go ahead and add two on both sides in order to fully isolate that c. Four plus two is six, so I get that six is equal to c. So now that I know what that constant is, I can go ahead and plug it back into that general solution to give me my particular solution. So y is then equal to negative two cosine x plus three sine x plus six. And this is my particular solution to my differential equation given the initial condition that y of zero is equal to four. Let us know if you have any questions, and I'll see you in the next one.

Find the particular solution to the differential equation y ′ = 2 sin ⁡ x + 3 cos ⁡ x y^{\prime}=2\sin x+3\cos x given the initial condition y ( 0 ) = 4 y\left(0\right)=4 .

y = − 2 cos ⁡ x + 3 sin ⁡ x + 6 y=-2\cos x+3\sin x+6

y = − 2 cos ⁡ x + 3 sin ⁡ x + 4 y=-2\cos x+3\sin x+4

y = − 2 cos ⁡ x + 3 sin ⁡ x + 1 y=-2\cos x+3\sin x+1

y = − 2 cos ⁡ x + 3 sin ⁡ x y=-2\cos x+3\sin x

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0. Functions7h 54m

Introduction to Functions
16m

Piecewise Functions
10m

Properties of Functions
9m

Common Functions
1h 8m

Transformations
5m

Combining Functions
27m

Exponent rules
32m

Exponential Functions
28m

Logarithmic Functions
24m

Properties of Logarithms
36m

Exponential & Logarithmic Equations
35m

Introduction to Trigonometric Functions
38m

Graphs of Trigonometric Functions
44m

Trigonometric Identities
47m

Inverse Trigonometric Functions
48m

1. Limits and Continuity2h 2m

Introduction to Limits
41m

Finding Limits Algebraically
40m

Continuity
40m

2. Intro to Derivatives1h 33m

Tangent Lines and Derivatives
30m

Derivatives as Functions
14m

Basic Graphing of the Derivative
23m

Differentiability
26m

3. Techniques of Differentiation3h 18m

Basic Rules of Differentiation
39m

Higher Order Derivatives
12m

Product and Quotient Rules
55m

Derivatives of Trig Functions
40m

The Chain Rule
49m

4. Applications of Derivatives2h 38m

Motion Analysis
36m

Implicit Differentiation
27m

Related Rates
59m

Linearization
13m

Differentials
21m

5. Graphical Applications of Derivatives6h 2m

Intro to Extrema
23m

Finding Global Extrema
50m

The First Derivative Test
1h 6m

Concavity
1h 1m

The Second Derivative Test
31m

Curve Sketching
44m

Applied Optimization
1h 25m

6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m

Derivatives of Exponential & Logarithmic Functions
1h 52m

Logarithmic Differentiation
20m

Derivatives of Inverse Trigonometric Functions
24m

7. Antiderivatives & Indefinite Integrals1h 26m

Antiderivatives
17m

Indefinite Integrals
25m

Integrals of Trig Functions
15m

Initial Value Problems
27m

8. Definite Integrals4h 44m

Estimating Area with Finite Sums
42m

Riemann Sums
1h 21m

Introduction to Definite Integrals
22m

Fundamental Theorem of Calculus
37m

Average Value of a Function
21m

Substitution
1h 19m

9. Graphical Applications of Integrals2h 27m

Area Between Curves
1h 18m

Introduction to Volume & Disk Method
1h 8m

10. Physics Applications of Integrals 3h 16m

Kinematics
1h 13m

Work
2h 3m

11. Integrals of Inverse, Exponential, & Logarithmic Functions2h 34m

Integrals of Exponential Functions
40m

Integrals Involving Logarithmic Functions
58m

Integrals Involving Inverse Trigonometric Functions
55m

12. Techniques of Integration7h 41m

Integration by Parts
2h 32m

Partial Fractions
4h 29m

Improper Integrals
39m

13. Intro to Differential Equations2h 55m

Basics of Differential Equations
48m

Slope Fields
19m

Euler's Method
22m

Separable Differential Equations
1h 25m

14. Sequences & Series5h 36m

Sequences
1h 22m

Review of Factorials
11m

Series
44m

Convergence Tests
3h 17m

15. Power Series2h 19m

Introduction to Power Series
1h 9m

Taylor Series & Taylor Polynomials
1h 10m

16. Parametric Equations & Polar Coordinates7h 58m

Parametric Equations
1h 6m

Calculus with Parametric Curves
1h 13m

Polar Coordinates
2h 5m

Calculus in Polar Coordinates
55m

Conic Sections
2h 36m

13. Intro to Differential Equations

Basics of Differential Equations

Calculus

13. Intro to Differential Equations

Basics of Differential Equations

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Multiple Choice

Find the particular solution to the differential equation $y^{'} = 2 \sin x + 3 \cos x$ given the initial condition $y of 0 equals 4$ .

$y = - 2 \cos x + 3 \sin x + 6$

$y = - 2 \cos x + 3 \sin x + 4$

$y = - 2 \cos x + 3 \sin x + 1$

$y = - 2 \cos x + 3 \sin x$

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Verified step by step guidance

Step 1: Recognize that the given differential equation y′=2sin(x)+3cos(x) is a first-order ordinary differential equation. To solve it, we integrate the right-hand side with respect to x to find y(x).

Step 2: Break the integration into two parts: ∫2sin(x)dx and ∫3cos(x)dx. Use the standard integration formulas: ∫sin(x)dx = -cos(x) and ∫cos(x)dx = sin(x).

Step 3: Perform the integrations: ∫2sin(x)dx = -2cos(x) and ∫3cos(x)dx = 3sin(x). Combine these results to get the general solution y(x) = -2cos(x) + 3sin(x) + C, where C is the constant of integration.

Step 4: Use the initial condition y(0) = 4 to solve for the constant C. Substitute x = 0 and y = 4 into the general solution: 4 = -2cos(0) + 3sin(0) + C. Simplify using cos(0) = 1 and sin(0) = 0 to find C.

Step 5: Substitute the value of C back into the general solution to obtain the particular solution y(x).