工程數學的問題請幫我解拜託了謝謝

Question

1.solve the initial value problem x dy/dx + y = e^x , y(1)=2
2.determine whether the differential equation 2xydx+(x^2-1)dy=0 is exact solve it
3.find the general solution of the equation y\'\'\'-y\'\'-y\'+y=0
4.solve the cauchy-euler equation x^2y\'\'-xy\'+10y=0
5.use the method of undetermined coefficients to solve the equation y\'\'+y\'-6y=2x
6.use u=y\' to solve yy\'\'=(y\')^2

Answer 1

1.solve the initial value problem x( dy／dx ) + y = ex , y(1) = 2sol：　　原式化為：( dy／dx ) + ( y／x ) = ex／x ～ 一階線性ODE　　積分因子：I(x) = e∫( 1／x )dx = eln│x│= x　　y = ( 1／x )[∫( x‧ex／x )dx + c ]　　   = ( 1／x )[ ex + c ] = ex／x + c／x　　代入邊界條件 x = 1 , y = 2　　→ e1 + c = 2 → c = 2 - e1　　→ y = ex／x + ( 2 - e1 )／x #2.determine whether the differential equation 2xydx + ( x2 - 1 )dy=0 is exact solve itsol：　　令 M( x , y ) = 2xy　　 　 N( x , y ) = x2 - 1　　∂M／∂y = 2x　,　∂N／∂x = 2x　　∂M／∂y = ∂N／∂x → exact　　若 Exact，則必有一解，其型式為：F( x , y ) = C　　F =∫Mdx + c( y )　　   =∫2xydx + c( y ) = x2y + c(y)　　∂F／∂y = N　　→ x2 + c'( y ) = x2 - 1　　→ c'( y ) = - 1 → c( y ) = - y　　所以解為：x2y - y = C #3. find the general solution of the equation y''' - y'' - y' + y = 0sol：　　特徵方程式：r3 - r2 - r + 1 = 0　　→ ( r - 1 )( r - 1 )( r + 1 ) = 0 → r = 1 , 1 , - 1　　→ y = ( c1 + c2x )ex + c3e-x #4. solve the cauchy - euler equation x2y'' - xy' + 10y = 0sol：　　令 y = xm 代入原式得：m( m - 1 ) - m + 10 = 0　　→ m2 - 2m + 10 = 0 → m = - 1 ± 3i　　→ y = x-1[ c1cos 3ln│x│+ c2sin 3ln│x│] #5.use the method of undetermined coefficients to solve the equation y'' + y' - 6y = 2xsol：　　特徵方程式：r2 + r - 6 = 0　　→ ( r - 2 )( r + 3 ) = 0 → r = 2 , - 3　　齊次解：yh = c1e2x + c2e-3x　　令 yp = Ax + B　　→ yp' = A　,　yp'' = 0　　yp'' + yp' - 6yp = 2x　　→ A - Ax - B = 2x　　比較係數得：- A = 2 → A = - 1／2　　　　　　　　A - B = 0 → B = - 1／2　　特解：yp = - x／2 - 1／2　　y = yh + yp　　→ y = c1e2x + c2e-3x - ( x／2 ) - ( 1／2 ) #6.use u = y' to solve yy'' = ( y' )2sol：　　令 u = y' → u' = yy''　　原式可寫成：u' = u2 ～ 變數可分離型ODE　　→∫( 1／u2 )du =∫dx + c1　　→ - 1／u = x + c1　　→ u = - 1／( x + c1 )　　→ y' = - 1／( x + c1 ) ～ 變數可分離型ODE　　→∫dy =∫[ - 1／( x + c1 ) ]dx + c2　　→ y = - ln│x + c1│+ c2 #　　希望以上解答的過程能幫這您。

Answer 2

ㄟˇ~~~~我覺得阿~~~
這些題目看起來都不難~~如果自己不去做做看就只是要找人來解題目的話說不太過去~~~
如果你是有其他原因才來問這些題目的解法的話,那就當我多嘴跟你說聲抱歉囉~~~