Differential Equations Exam 2 Study Guide and Practice Problems.pdf Differential Equations Exam 2 Study Guide and Practice Problems.pdf Differential Equations Exam 2 Study Guide and Practice Problems.pdf
Differential Equations
Exam 2 Study Guide and
Practice Problems
Guidehttps://www.stuvia.com/dashboard!@_)#*)(@$)($@*($@)($@*_
Differential Equations Exam 2 Study Guide and Practice Problems.pdf Differential Equations Exam 2 Study Guide and Practice Problems.pdf Differential Equations Exam 2 Study Guide and Practice Problems.pdf
, Differential Equations Exam #2.pdf Differential Equations Exam #2.pdf Differential Equations Exam #2.pdf
Exact Equation M(x,y)dx + N(x,y)dy = 0 where fx(x,y)= M(x,y) and fy(x,y)=N(x,y) with
solution f(x,y)=C
M(x,y)dx + N(x,y)dy = 0 is exact if My(x,y)=Nx(x,y)
if the equation is exact, solve by integrating f(x,y)=∫M(x,y)dx + g(y) where
g(y) is an arbitrary integration function, then differentiate f(x,y) to get δf/
δy=δ/δy∫M(x,y)dx + g'(y)= N(x,y) and solve for g'(y)
recall: δf/δy= N(x,y)
Integrating Factor Exact Equations in some cases, multiplying by µ(x,y) makes an equation exact
If (My-Nx)/N is a function of x only (no y's) then µ=e^∫[(My-Nx)/N]dx
If (Nx-My)/M is a function of y only (no x's) then µ=e^∫[(Nx-My)/M]dx
µ(x,y)M(x,y)dx + µ(x,y)N(x,y)dy = 0 is an exact equation
Substitution - to simplify to DE set y=g(x,u) and transform dy/dx=f(x,y) into
du/dx=F(x,u)
- define u and then solve for du/dx
Differential Equations Exam #2.pdf Differential Equations Exam #2.pdf Differential Equations Exam #2.pdf
Differential Equations
Exam 2 Study Guide and
Practice Problems
Guidehttps://www.stuvia.com/dashboard!@_)#*)(@$)($@*($@)($@*_
Differential Equations Exam 2 Study Guide and Practice Problems.pdf Differential Equations Exam 2 Study Guide and Practice Problems.pdf Differential Equations Exam 2 Study Guide and Practice Problems.pdf
, Differential Equations Exam #2.pdf Differential Equations Exam #2.pdf Differential Equations Exam #2.pdf
Exact Equation M(x,y)dx + N(x,y)dy = 0 where fx(x,y)= M(x,y) and fy(x,y)=N(x,y) with
solution f(x,y)=C
M(x,y)dx + N(x,y)dy = 0 is exact if My(x,y)=Nx(x,y)
if the equation is exact, solve by integrating f(x,y)=∫M(x,y)dx + g(y) where
g(y) is an arbitrary integration function, then differentiate f(x,y) to get δf/
δy=δ/δy∫M(x,y)dx + g'(y)= N(x,y) and solve for g'(y)
recall: δf/δy= N(x,y)
Integrating Factor Exact Equations in some cases, multiplying by µ(x,y) makes an equation exact
If (My-Nx)/N is a function of x only (no y's) then µ=e^∫[(My-Nx)/N]dx
If (Nx-My)/M is a function of y only (no x's) then µ=e^∫[(Nx-My)/M]dx
µ(x,y)M(x,y)dx + µ(x,y)N(x,y)dy = 0 is an exact equation
Substitution - to simplify to DE set y=g(x,u) and transform dy/dx=f(x,y) into
du/dx=F(x,u)
- define u and then solve for du/dx
Differential Equations Exam #2.pdf Differential Equations Exam #2.pdf Differential Equations Exam #2.pdf
