Find the function F that satisfies the following differential equations and initial conditions.F″(x)=cos⁡x,F′(0)=3,F(π)=5

Name: Find the function F that satisfies the following differential equations and initial conditions.F″(x)=cos⁡x,F′(0)=3,F(π)=5
Uploaded: 2022-02-23T17:22:57+00:00
Description: Find the function F that satisfies the following differential equations and initial conditions.F″(x)=cos⁡x,F′(0)=3,F(π)=5

Question

Cassandra Ramirez · Accepted Answer

Step 1Given that,(1) f″(x)=cos⁡x(2) f′(0)=3(3) f(π)=5Integrate equation 1;∫fx)dx=∫cos⁡(x)dxf′(x)=sin⁡(x)=sin⁡(x)+c (4)Using criteria (2);f′(0)=sin⁡(0)+c3=0+c⇒c=3Equation (4) is are written as,f′(x)=sin⁡(x)+3 ..(5)Step 2Integrate equation 5;∫f′(x)dx=∫[sin⁡(x)+3]dxf(x)=−cos⁡(x)+3x+c′ (5)Using criteria (3),f(π)=−cos⁡(π)+3π+c′5=1+3π+c′⇒c′=4−3πEquation (5) is are written as,f(x)=−cos⁡(x)+3x+4−3π

Nadine Salcido · Accepted Answer

As we know Fx)=ddxF′(x). Therefore, F&#039;(x) is an antiderivative of F&quot;(x).F′(x)=∫Fx)dx=∫cos⁡xdx=sin⁡x+C1 (2)Taking into account that F′(0)=3 we have:F′(0)=3⇒sin⁡(0)+C1=3⇒C1=3 (3)Then F′(x)=sin⁡x+3Since F′(x)=ddxF(x) we have that:F(x)=∫F′(x)dx=∫(sin⁡x+3)dx=∫sin⁡xdx+∫3dx (4)=−cos⁡x+3x+C2Since F(π)=4 we have thatF(π)=4⇒−cos⁡(π)+3(π)+C2=4⇒1+3π+C2=4 (5)⇒C2=3(1−π)Finally, we have found thatF(x)=−cos⁡x+3x+3(1−π) (6)

karton · Accepted Answer

From the given conditions we haveF′(x)=∫cos⁡xdx=sin⁡x+CNow if we include that F′(0)=3, we have3=F′(0)=Cthat is, C=3, which means that F′(x)=sin⁡x+3Now we have thatF(x)=∫(sin⁡x+3)dx=∫sin⁡xdx+∫3dx=−cos⁡x+3x+DFrom the given condition we get4=F(π)=1+3π+Dthat is, D=3−3π, so F(x)=−cos⁡x+3x+3−3π

Find the function F that satisfies the following differential equations

Answered question

Answer & Explanation

New Questions in Differential Equations