227k views
2 votes
Solve the differential equation by variation of parameters. y'' + y = cos2(x)

User Chris Yo
by
7.6k points

1 Answer

3 votes

Answer:


y=c_1\cos(x)+c_2\+\sin(x)+\sin^2(x)-(1)/(3)\sin^4(x)+(1)/(3)\cos^4(x)}}}

Explanation:

Given the second-order differential equation,
y'' + y = cos2(x), solve it using variation of parameters.

(1) - Solve the DE as if it were homogenous and find the homogeneous solution
y'' + y = cos2(x) \Longrightarrow y'' + y =0\\\\\text{The characteristic equation} \Rightarrow m^2+1=0\\\\m^2+1=0\\\\ \Longrightarrow m^2=-1\\\\\ \Longrightarrow m=√(-1) \\\\\Longrightarrow \boxed{m=\pm i} \\ \\\text{Solution is complex will be in the form} \ \boxed{y=c_1e^(\alpha t)\cos(\beta t)+c_2e^(\alpha t)\sin(\beta t)} \ \text{where} \ m=\alpha \pm \beta i


\therefore \text{homogeneous solution} \rightarrow \boxed{y_h=c_1\cos(x)+c_2\sin(x)}

(2) - Find the Wronskian determinant


|W|=\left|\begin{array}{ccc}y_1&y_2\\y'_1&y'_2\end{array}\right| \\\\\Longrightarrow |W|=\left|\begin{array}{ccc}\cos(x)&\sin(x)\\-sin(t)&cos(x)\end{array}\right|\\\\\Longrightarrow \cos^2(x)+\sin^2(x)\\\\\Longrightarrow \boxed=1

(3) - Find W_1 and W_2


\boxed{W_1=\left|\begin{array}{ccc}0&y_2\\g(x)&y'_2\end{array}\right| and \ W_2=\left|\begin{array}{ccc}y_2&0\\y'_2&g(x)\end{array}\right|}


W_1=\left|\begin{array}{ccc}0&\sin(x)\\\cos^2(x)&\cos(x)\end{array}\right|\\\\\Longrightarrow \boxed{W_1= -\sin(x)\cos^2(x)}\\\\W_2=\left|\begin{array}{ccc}\cos(x)&0\\ -\sin(x)&\cos^2(x)\end{array}\right|\\\\\Longrightarrow \boxed{W_2= \cos^3(x)}

(4) - Find u_1 and u_2


\boxedW\

u_1:


\int((-\sin(x)\cos^2(x))/(1)) dx\\\\\Longrightarrow-\int(\sin(x)\cos^2(x)) dx\\\\\text{Let} \ u=\cos(x) \rightarrow du=-sin(x)dx\\\\\Longrightarrow\int u^2 du\\\\\Longrightarrow (1)/(3)u^3\\ \\\Longrightarrow \boxed{u_1=(1)/(3)\cos^3(x)}

u_2:


\int(\cos^3(x))/(1)dx\\ \\\Longrightarrow \int \cos^3(x)dx\\\\ \Longrightarrow \int (\cos^2(x)\cos(x))dx \ \ \boxed{\text{Trig identity:} \cos^2(x)=1-\sin^2(x)}\\\\\Longrightarrow \int[(1-\sin^2(x)})\cos(x)]dx\\\\\Longrightarrow \int \cos(x)dx-\int (\sin^2(x)\cos(x))dx\\\\\Longrightarrow \sin(x)-\int (\sin^2(x)\cos(x))dx\\\\\text{Let} \ u=\sin(x) \rightarrow du=cos(x)dx\\\\\Longrightarrow \sin(x)-\int u^2du\\\\\Longrightarrow \sin(x)-(1)/(3) u^3\


\Longrightarrow \boxed{u_2=\sin(x)-(1)/(3) \sin^3(x)}

(5) - Generate the particular solution


\text{Particular solution} \rightarrow y_p=u_1y_1+u_2y_2


\Longrightarrow y_p=((1)/(3)\cos(x))(\cos(x))+(\sin(x)-(1)/(3) \sin^3(x))(\sin(x))\\\\ \Longrightarrow y_p=(1)/(3)\cos^4(x)+\sin^2(x)-(1)/(3)\sin^4(x)\\\\\Longrightarrow \boxed{y_p=\sin^2(x)-(1)/(3)\sin^4(x)+(1)/(3)\cos^4(x)}

(6) - Form the general solution


\text{General solution} \rightarrow y_(gen.)=y_h+y_p


\boxed{\boxed{y=c_1\cos(x)+c_2\+\sin(x)+\sin^2(x)-(1)/(3)\sin^4(x)+(1)/(3)\cos^4(x)}}}

Thus, the solution to the given DE is found where c_1 and c_2 are arbitrary constants that can be solved for given an initial condition. You can simplify the solution more if need be.

User Brian Trzupek
by
7.9k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories