solution of the differential equation 3 y2 dy + 2x = 1 dx giving the answer in the form y 3 = f ( x ) . C4A , y 3 = A − x 2 + x Question 2 (**) Find a general solution of the differential equation dy = xy , x ≠ 0 , y ≠ 0 , dx giving the answer in the form y = f ( x ) . 1 x2 y = Ae 2 Created by T. Madas
solution of the differential equation dy = ( y + 1)(1 − 2 x ) , y ≠ −1 . dx giving the answer in the form y = f ( x ) . 2 y = A e x− x − 1 Question 4 (**+) Find a general solution of the differential equation dy = y tan x , y > 0 dx giving the answer in the form y = f ( x ) . y = A sec x Created by T. Madas
solution of the differential equation dy = 2e x − y dx giving the answer in the form y = f ( x ) . ( y = ln 2 e x + C ) Question 6 (***) Find a general solution of the differential equation x2 dy = xy + y , x ≠ 0 , y ≠ 0 , dx giving the answer in the form y = f ( x ) . [the final answer may not contain natural logarithms] −1 C4A , y = Ax e x Created by T. Madas
solution of the differential equation = xy , y > 0 , ( x2 + 3) dy dx giving the answer in the form y 2 = f ( x ) . ( y 2 = A x2 + 3 ) Question 8 (***+) Show that a general solution of the differential equation 2 dy y = , dx x is given by y= x , 1 + Ax where A is an arbitrary constant. proof Created by T. Madas
solution of the differential equation dy x ex = , dx sin y cos y giving the answer in the form f ( x, y ) = constant . cos 2 y + 4e x ( x − 1) = C or e x ( x − 1) − sin 2 y = C or e x ( x − 1) + cos 2 y = C Question 10 (***+) Find a general solution of the differential equation dy cos 2 x = y 2 sin 2 x dx giving the answer in the form y = f ( x ) . y= Created by T. Madas 1 C + x − tan x
solution of the differential equation sec3 x dy = cot 2 2 y dx giving the answer in the form f ( x, y ) = c . 3tan 2 y − 6 y − 2sin 3 x = C Question 12 (***+) Find a general solution of the differential equation e2 x dy = cosec2 y dx giving the answer in the form f ( x, y ) = c . 2 x + 2e −2 x − sin 2 y = C Created by T. Madas
general solution of the differential equation 5 dy = 2 y2 − 7 y + 3 dx is given by y= A ex − 3 2 A ex −1 , where A is an arbitrary constant. SYN-A , proof Created by T. Madas
general solution of the differential equation e x+2 y dy 2 + (1 − x ) = 0 dx is given by 1 y = ln 2 e− x x 2 + 1 + K , 2 ( ) where K is an arbitrary constant. proof Created by T. Madas
x− y+3, x > 0. dx Determine a general solution of the above differential equation, by using the substitution u = y x . −1 SP-W , y = 1 x + 3 + Cx 2 3 Created by T. Madas
substitution y = xu , where u = f ( x ) , or otherwise, find a simplified general solution for the following differential equation. x dy = 2 x 2 + 2 xy + y . dx SP-J , y = Ax e− x − x Created by T. Madas
find a simplified general solution for the following differential equation. ( 2 dy dy x − 1 − 2 xy + y 2 = 1 . dx dx 2 ) SP-F , ( A + y x −1) ( y + Bx + C ) = 0 Created by T. Madas 2
the x axis at the origin O . It is further given that the equation of such a circle satisfies the differential equation = y f ( x) , ( x2 − y 2 ) dy dx for some function f . Use an algebraic method to find an expression for f ( x ) . SP-W , f ( x ) = 2 x Created by T. Madas
functions u ( x ) and v ( x ) satisfy the integral equations u ( x ) dx = ux 2 and u ( x ) v ( x ) dx = u ( x ) dx v ( x ) dx . Determine, in terms of an arbitrary constant, a simplified expression for u ( x ) and a 2 similar expression for v ( x ) . −1 Ae x 2 B SP-L , u ( x ) = 2 , v ( x ) = 2 x (1 − x ) 1 − x 2 ( Created by T. Madas )
of the quadratic equation x 2 − x − 1 = 0 is denoted by φ , and is commonly known as the golden section or golden number. a) Show, with a detailed method, that F ( x ) = f (φ ) x ( ) is a solution of the g φ differential equation, F ′ ( x ) = F −1 ( x ) , where f and g are constant expressions of φ , to be found in simplified form. b) Verify the answer obtained in part (a) satisfies the differential equation, by differentiation and function inversion. [You may assume that F ( x ) is differentiable and invertible] 1 1 φ SPX-L , F ( x ) = xφ = φ1−φ xφ φ Created by T. Madas
equation dy 4 x + = 0, y ≠ 0, dx y subject to the condition y = 2 at x = 0 . Give the answer in the form f ( x, y ) = constant. 4 x2 + y 2 = 4 Question 2 (**) Solve the differential equation dy cos 2 x = , dx y subject to the condition y = 6 at x = π 4 x > 0, y > 0 , , giving the answer in the form y 2 = f ( x ) . y 2 = 35 + sin 2 x Created by T. Madas
equation dy = 6 xy 2 , dx with y = 1 at x = 2 , giving the answer in the form y = f ( x ) . y= 1 13 − 3x 2 Question 4 (**) Solve the differential equation dy 3sin 3 x = , dx y subject to the condition y = 3 at x = π 3 , giving the answer in the form y 2 = f ( x ) . y 2 = 7 − 2 cos 3 x Created by T. Madas
equation dy 2 x = , dx y with y = 2 at x = 1 , giving the answer in the form y 2 = f ( x ) . y 2 = 2 x2 + 2 Question 6 (**) Solve the differential equation dy 10 , = dx ( x + 1)( x + 2 ) subject to the condition y = 0 at x = 0 , giving the answer in the form y = f ( x ) . y = 10ln Created by T. Madas 2x + 2 x+2
equation ( ) 1 dy cos 3 x = , dx y subject to the condition y = 1 at x = π 2 , giving the answer in the form y 2 = f ( x ) . ( ) y 2 = 6sin 1 x − 2 3 Question 8 (**) Solve the differential equation dy = 3x 2 y dx subject to the condition y = 0 at x = 1 , giving the answer in the form y = f ( x ) . ( ) y = 1 x3 − 1 4 Created by T. Madas 2
equation dy = dx y , x +1 y ≠ 0 , x ≠ −1 , subject to the condition y = 9 at x = 8 , giving the answer in the form y = f ( x ) . y = x +1 Question 10 (**) Solve the differential equation dy = 2 x 2 y − 1, dx y>1 2 subject to the condition y = 1 at x = 0 , giving the answer in the form y = f ( x ) . 2 ( ) y = 1 x4 + 1 2 Created by T. Madas
equation dy + y 2 e x = 0, dx subject to the condition y = 1 at x = 0 , giving the answer in the form y = f ( x ) . 2 y= 1 x e +1 Question 12 (**) dy 2 y = , x > 0 , y > 0. dx x Show that the solution of the above differential equation subject to the boundary condition y = 3 at x = 1 , is given by y = 3x 2 . proof Created by T. Madas
equation dy = yx 2 , x ≠ 0 , y ≠ 0 , dx subject to the condition y = 1 at x = 1 , giving the answer in the form y = f ( x ) . ( ) 1 x3 −1 y = e3 Created by T. Madas
equation dy = y2 x , x ≠ 0 , y ≠ 0 , dx with y = −2 at x = 1 . Give the answer in the form y = A 3 , where A and B are integers. 1 + Bx 2 SYN-N , y = 6 3 1− 4x 2 Created by T. Madas
equation x3 dy = 2 y2 dx subject to the condition y = 1 at x = 1 , giving the answer in the form y = f ( x ) . 2 y= x2 x2 + 1 Question 16 (**+) Solve the differential equation dy x − y +e =0, dx subject to y = 0 at x = 0 , giving the answer in the form f ( x, y ) = constant . ex + e y = 2 Created by T. Madas
equation dy = xy e x , x > 0 , y > 0 , dx subject to boundary condition y = e at x = 1 . Give the answer in the form ln y = f ( x ) . ln y = x e x − e x + 1 Question 18 (**+) Solve the differential equation 4 y +1 dy =− dx x2 subject to the condition, y = 2 at x = 2 , giving the answer in the form y = f ( x ) . 3 y= Created by T. Madas 1 x 2 − 1 4
= 2 at x = 0 , solve the differential equation dy = 4 + y2 , dx giving the answer in the form y = f ( x ) . You may assume that 1 2 a +x 2 dx = 1 x arctan + C . a a π y = 2 tan 2 x + 4 Question 21 (***) ( x + 1) dy = 3y , y > 0 . dx Solve the differential equation subject to the condition y = 16 at x = 1 , to show that 3 y = 2 ( x + 1) . proof Created by T. Madas
equation dy y = , y>0 dx x ( 2 − x ) subject to the condition y = 1 at x = 1 , giving the answer in the form y 2 = f ( x ) . C4M , y 2 = x 2− x Question 23 (***) Find the solution of the differential equation dy = y sin x , y > 0 dx subject to the condition y = 10 at x = π , giving the answer in the form y = f ( x ) . y = 10e−1−cos x Created by T. Madas
equation x2 dy = y 2 − 3x 4 y 2 dx subject to the condition y = 1 at x = 1 , giving the answer in the form y = f ( x ) . 2 C4K , y = x 4 x +1 Question 25 (***) Solve the differential equation dy = 4 yx3 , dx y≠0 subject to the condition y = 1 at x = 1 , giving the answer in the form y = f ( x ) . 4 y = e x −1 Created by T. Madas
equation = y ( x + 1) , y ≠ 0, x ≠ ±1 , (1 − x2 ) dy dx subject to the condition y = 2 at x = 1 , giving the answer in the form y = f ( x ) . 2 y= 1 1− x Question 27 (***) Solve the differential equation dy 2 x ln x = , x >0, y >0 dx y subject to the condition y = 2 e at x = e , giving the answer in the form y 2 = f ( x ) . y 2 = x 2 ( 2ln x − 1) + 3e2 Created by T. Madas
equation dy = 4 xy − 3 yx 2 dx subject to the condition y = 1 at x = 2 , giving the answer in the form y = f ( x ) . 2 SYN-I , y = e 2 x − x Question 29 3 (***) Solve the differential equation dy 10 − y = dx 5 subject to the condition y = 1 at x = 0 , giving the answer in the form y = f ( x ) . −1x y = 10 − 9e 5 Created by T. Madas
solution of the differential equation 2 y −1 dy = , dx x2 x ≠ 0, y > 1 2 subject to the condition y = 1 at x = 1 , is given by y= 5x2 − 4 x + 1 2 x2 . proof Created by T. Madas
equation x ( x + 2) dy = y, x > 0, y > 0 dx subject to the condition y = 2 at x = 2 , giving the answer in the form y 2 = f ( x ) . MP2-N , y 2 = Created by T. Madas 8x x+2
equation dy 5y = dx ( 2 + x )(1 − 2 x ) subject to the condition y = 2 at x = 0 , giving the answer in the form y = f ( x ) . C4E , y = 2+ x 1 − 2x Question 33 (***+) Solve the differential equation dy y = , y > 0 , x > −1 dx ( x + 1)( x + 3) subject to the condition y = 2 at x = 1 , giving the answer in the form y 2 = f ( x ) . y2 = Created by T. Madas 8 ( x + 1) x+3
2 )( x + 3) dy = 7 y , y > 0 , x > −3 . dx Show that the solution of the above differential equation subject to the boundary condition, y = 6 at x = 4 , is given by y= 3 ( 3x + 2 ) . x+3 proof Question 35 (***+) Solve the differential equation ey dy + x ex = 0 , x < 1 dx subject to the condition y = 0 at x = 0 , giving the answer in the form y = f ( x ) . y = x + ln (1 − x ) Created by T. Madas
equation ( 2 x − 3)( x − 1) dy = y ( 2 x − 1) , y ≠ 0 , dx subject to the condition y = 10 at x = 2 , giving the answer in the form y = f ( x ) . 10 ( 2 x − 3) C4I , y = x −1 2 Question 37 (***+) Solve the differential equation ex subject to y = π 4 dy x = , 0< y <π , dx sin 2 y at x = 0 , giving the answer in the form cos f ( y ) = g ( x ) . cos 2 y = 2 x e− x + 2e − x − 2 Created by T. Madas
equation y (13 − 2 x ) dy = , y ≠0, dx ( 2 x − 3)( x + 1) subject to the condition y = 4 at x = 2 , giving the answer in the form y = f ( x ) . y= 108 ( 2 x − 3) 2 ( x + 1)3 Question 39 (***+) Solve the differential equation dy = yx 2 cos x, x > 0, y > 0 dx subject to the condition y = 1 at x = π , giving the answer in the form ln y = f ( x ) . ln y = x 2 sin x + 2 x cos x − 2sin x + 2π Created by T. Madas
= , y ≠0, dx x + 3 Show that the solution of the above differential equation, subject to the boundary condition y = 1 at x = 1 , can be written as y 2 = ln e ( x + 3) . 4 proof Question 41 (***+) dy cos 2 4 x = y , y > 0 . dx Show that the solution of the above differential equation subject to the boundary condition y = e3 at x = π 16 is given by 1 (11+ tan 4 x ) y = e4 . C4G , proof Created by T. Madas
y = a at t = 0 , the solution of the differential equation 1 dy = ω a2 − y2 2 , dt ( ) where a and ω are positive constants, can be written as y = a cos ωt . You may assume that x dx = arcsin + C . a a2 − x2 1 proof Question 43 (***+) Solve the differential equation yey 2 dy dx = e2 x , x ≠ 0 , y ≠ 0 , subject to the condition y = 2 at x = 2 , giving the answer in the form y 2 = f ( x ) . y2 = 2x Created by T. Madas
− 2 )( y + 1) ≡ P Q + , y ≠ −1, 2 ( y − 2 ) ( y + 1) a) Find the value of each of the constants P and Q . b) Hence, show that the solution of the differential equation dy = x 2 ( y − 2 )( y + 1) dx can be written as 3 y−2 = A e x , where A is a constant. y +1 c) Given further that y = 5 when x = 0 , show clearly that y= 4 + ex 3 2 − ex 3 . P = 1, Q = −1 3 3 Created by T. Madas
equation dy y = , dx ( 2 x + 1)( x + 1) 1 x>− , y>0 2 subject to the condition y = 2 at x = 0 , giving the answer in the form y = f ( x ) . y= 4x + 2 x +1 Question 46 (***+) Solve the differential equation dy y = , dx x ( x + 1)2 subject to the condition y = x > 0, y > 0 1 at x = 1 , giving the answer in the form ln y = f ( x ) . 2 1 1 x ln y = ln − + x +1 x +1 2 Created by T. Madas
equation dy = 3 x e3 x + y dx subject to the condition y = 0 at x = 0 , giving the answer in the form e y = f ( x ) . ey = 3 3x e − 3 x e3 x + 2 Question 48 (***+) −5 dy = 2 y − 150 , y < 75 dx Solve the above differential equation, given that when x = 0 , y = 275 . Give the answer in the form y = f ( x ) . −2x C4B , y = 75 + 200e 5 Created by T. Madas
) dydx = x (1 + y ) , with y = 0 at x = 0 . Show that the solution of the above differential equation is ( ) 1 y = 1 + x2 2 − 1 . proof Created by T. Madas
y ( y + 1) , x > 0 , y > 0 dx Show that the solution of the above differential equation subject to the boundary condition y = 1 at x = 1 , is given by 2 3 y= x . 1− x MP2-L , proof Created by T. Madas
y 1 + 2 x2 , x > 0 , y > 0 dx ( ) Show that the solution of the above differential equation subject to y = e at x = 1 , is 2 y = x ex . proof Created by T. Madas
− 1 = , x >0, y >0 dx x Show that the solution of the above differential equation subject to y = 2 at x = 1 , is y= 3 + x2 3 − x2 . SYN-E , proof Created by T. Madas
y 2 = xy 2 , x > 0 , y > 0 dx Show that the solution of the above differential equation subject to y = e at x = 1 , is y= 1 x e . x MP2-B , proof Created by T. Madas
equation dy = x 2 e3 y − x dx subject to the condition y = 0 at x = 0 , giving the answer in the form e ( ) = g ( x ) . f y ( ) e −3 y = 3e − x x 2 + 2 x + 2 − 5 Question 55 (****) Solve the differential equation 5 1 dy = 2 x 2 + 1 cos 2 2 y x dx ( subject to x = 0 , y = π 8 ) , giving the answer in the form tan f ( y ) = g ( x ) 6 tan 2 y = 1 2 x 2 + 1 + 11 12 ( Created by T. Madas )
equation ( 2 x + 1) − dy ( 2 x − 1)3 = 0 dx subject to the condition y = 0 at x = 0 , giving the answer as y = f ( x ) . y=− Created by T. Madas x ( 2 x − 1)2
2 , x, y > 2 dx x − 3 x + 2 Solve the differential equation above, subject to the boundary condition y = 1 at 3 x = 3 , to show that 2 2 ( x − 2) y= . 3 ( x − 1) proof Created by T. Madas
equation dy = x sin 2 x cos 2 y dx subject to the condition x = π 4 , y = 0 , giving the answer in the form tan y = f ( x ) . tan y = 1 ( sin 2 x − 2 x cos 2 x − 1) 4 Question 59 (****) ( x − 1) dy = 2x y dx Solve the differential equation above, subject to the boundary condition y = 4 at x = 2 , to show that e y = e x ( x − 1) . proof Created by T. Madas
= y 2 − y dx Solve the differential equation above, subject to the boundary condition, y = 1 at 2 x = 0 , to show that y= 1 1 + esin x . MP2-E , proof Created by T. Madas
equation y (1 − x ) dy = , dx (1 + x ) 1 + x 2 ( ) subject to the condition x = 0 , y = 1 , giving the answer in the form y 2 = f ( x ) . 2 1+ x) ( y = 2 1 + x2 Question 62 (****) Solve the differential equation (1 + x2 ) dydx = x (1 − y 2 ) , y ≠ ±1 , subject to the condition y = 0 at x = 0 , giving the answer in the form y = f ( x ) . y= Created by T. Madas x2 x2 + 2
) dy = y (1 − x ) , y > 0 , x > −1 . dx Solve the above given differential equation, subject to the boundary condition y = 1 at x = 0 , to show that y = ( x + 1) e − x . 2 MP2-Z , proof Created by T. Madas
equation dy = 24cos 2 y cos3 x dx subject to the condition y = π 4 at x = π 6 , giving the answer in the form tan y = f ( x ) . tan y = 24sin x − 8sin 3 x − 10 Created by T. Madas
equation 2 dy = 2− 2 , dx y subject to the condition y = 2 at x = 1 , giving the answer in the form x = f ( y ) . 3y − 3 x = 1 y + 1 ln 2 4 y +1 Created by T. Madas
+ x ) dy = y. dx Solve the differential equation subject to the condition y = 3 at x = 0 , to show that y = 3 (1 + x ) e − x . 2 proof Created by T. Madas
= 1 − y 2 . dx Solve the differential equation above, subject to the boundary condition y = 0 at x= π 4 , to show that y= 1 − 2cos 2 x 1 + 2cos 2 x . SYN-H , proof Created by T. Madas
2) dy + y ( x + 1) = 0 , x > −2 . dx Solve the differential equation above, subject to the boundary condition y = 2 at x = 0 , to show that y = ( x + 2 ) e− x . proof Created by T. Madas
= f ( x ) satisfies the differential equation y = 1− dy x +1 , y > 1 , x > −1 dx ( x − 1)( x + 2 ) a) Solve the differential equation to show that ln ( y − 5 ) + 1 2 x + 4 x − 2ln ( x + 1) = C . 2 When x = 0 , y = 2 . b) Show further that 2 2 −1 x y = 1 + ( x + 1) e 2 . proof Created by T. Madas
equation 50 dy = 20 − y , dx given that when x = 0 , y = 0 , giving the answer in the form x = f ( y ) . x = 2000 ln Created by T. Madas 20 − 100 y 20 − y
equation d2y dx 2 +2 dy = 1, dx dy given that y = − 1 and = 1 at x = 0 , giving the answer in the form y = f ( x ) . 4 dx MP2-W , y = 1 2 x − e−2 x 2 Created by T. Madas
= f ( x ) satisfies the differential equation dy k ( 9 − x ) = , y > 0, 0 ≤ x ≤ 9 , dx y where k is a positive constant. dy It is further given that y = 1 , = 2 at x = 1 . 2 dx Find the possible values of x when dy 1 = . dx 5 SYN-V , x = 5 ∪ x = 13 Created by T. Madas
through the point with coordinates 1 , log 2 ( log 2 e ) and its gradient function satisfies dy = 2y , x ∈ , x < 2 . dx Find the equation of the curve in the form y = f ( x ) X , y = − log 2 ( 2 − x ) ln 2 Created by T. Madas
= f ( x ) satisfies the differential equation 2 xy ( y + 1) dy = , dx sin 2 x + 1 π 6 ( ) subject to the condition y = 1 at x = 0 . Find the exact value of y when x = π 12 . 1 MP2-S , y = e Created by T. Madas 1π 6 −1
form y = f ( x ) , a simplified solution for the following differential equation. dy cos x + 4 y 2 sin x = sin x , dx y = 15 at x = 1 π . 34 3 SP-H , y = Created by T. Madas sec4 x − 1 ( ) 2 sec 4 x + 1
= f ( t ) satisfies the differential equation dv 1 1 = k − , dt v h where k and h are non zero constants. Given that v = h − 1 at t = 0 , solve the differential equation to show that k t +v (h − v)eh =A where A is a non zero constant. SYN-X , proof Created by T. Madas
= f ( x ) satisfies the differential equation dy 1 1 = k − , dx h x where k and h are non zero constants. It is further given that y = −1 , dy d2y = −1 , = 2 at x = −1 . dx dx 2 Solve the differential equation to show that 2 e y−x = ( y + 2) . SPX-F , proof Created by T. Madas
= f ( x ) satisfies the differential equation d dy d 2 yx 2 = x , x>0 dx dx dx ( ) ( ) subject to the condition y = 4 at x = 3 . Find a simplified expression for y = f ( x ) . SP-I , y = Created by T. Madas 4 ( x − 2 )2
equation d2y 2 dy + 4 = 1, 2 dx dx given that y = 0 and dy 1 = at x = 0 , giving the answer in the form y = f ( x ) . dx 6 1 1 + 2 e 4 x 1 SP-T , y = ln − x 4 3 2 Created by T. Madas
differential equation d2y 2 dy = x , 2 dx dx given further y = 0 , dy = 2 at x = 0 . dx Give the answer in the form x = f ( y ) . SP-N , x = Created by T. Madas e y −1 ( ) = tanh 1 x 2 e +1 y
y4 − y2 x4 − x2 , x > 0 , y > 0. Find the solution of the above differential equation subject to the boundary condition 2 y= at x = 2 . 3 Give the answer in the form y = 2x , where f ( x ) is a function to be found. f ( x) MP2-T , f ( x ) = 3 + x 2 − 1 Created by T. Madas
function f ( x ) satisfies the integral equation f ( x ) dx = f ( x ) dx , f (0) = 1 . 4 2 dy Use the substitution f ( x ) = , to find a simplified expression for f ( x ) . dx SP-K , f ( x ) = 1 e 4 x 4 Created by T. Madas
to sketch the curve with equation y = f ( x ) , defined over the set of real numbers, in the greatest domain. The curve has the property that at every point on the curve, the second derivative equals to the first derivative squared. Showing all the relevant details, sketch the graph of y = f ( x ) , given further that the curve passes through the point ( 0, 2 ) and the gradient at that point is 1. SPX-O , graph Created by T. Madas
that a function with equation y = f ( x ) is a solution of the following differential equation. 2 (1 − x2 ) ddx2y − 2 x dydx + y = 0 . Show with a clear method that x 1 ( ) f ( x ) dx = x 2 − 1 f ′ ( x ) . SP-X , proof Created by T. Madas
differential equation 2 y d2y dy dy dy + + 2 y = 0 , y ( 0) = 2 , ( 0 ) = − 12 . 2 dx dx dx dx Give the answer in the form y 2 = f ( x ) . SPX-G , y 2 = 3 + e−2 x Created by T. Madas
equation y = f ( x ) satisfies the differential equation d2y dx 2 = 2 dy 1 − , 2 x − 1 dx y ( 0) = 1, dy ( 0 ) = −1 . dx Solve the above differential equation giving the answer in the form y = f ( x ) . SPX-B , y = x + 1 + ln 2 x − 1 Created by T. Madas
of the quadratic equation x 2 − x − 1 = 0 is denoted by φ , and is commonly known as the golden section or golden number. a) Show, with a detailed method, that F ( x ) = f (φ ) x ( ) is a solution of the g φ differential equation, F ′ ( x ) = F −1 ( x ) , where f and g are constant expressions of φ , to be found in simplified form. b) Verify the answer obtained in part (a) satisfies the differential equation, by differentiation and function inversion. [You may assume that F ( x ) is differentiable and invertible] 1 1 φ V , SPX-L , F ( x ) = xφ = φ1−φ xφ φ Created by T. Madas
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