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

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 $y of 0 equals 4$ .

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

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

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

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

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).

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