Question 1

Passage: Consider the differential equation e^ x+y dy dx =e^ x-y . Question: What is d^2y dx^2 ( dx dy )^2 equal to ?

Accepted Answer

Given the differential equation: e^ x+y dy dx = e^ x-y e^x e^y dy dx = e^x e^ -y Dividing both sides by e^x: dy dx = e^ -y e^y = e^ -2y Differentiating both sides with respect to x: d^2y dx^2 = d dx (e^ -2y ) = -2e^ -2y dy dx Substituting dy dx = e^ -2y : d^2y dx^2 = -2e^ -2y (e^ -2y ) = -2e^ -4y From dy dx = e^ -2y , we get: dx dy = 1 e^ -2y = e^ 2y Squaring both sides: ( dx dy )^2 = (e^ 2y )^2 = e^ 4y Now, multiplying the two results: d^2y dx^2 ( dx dy )^2 = (-2e^ -4y )(e^ 4y ) = -2e^0 = -2 An

Question 2

What is the solution of the differential equation ( dy dx )=p when y(0)=q ?

Accepted Answer

The given differential equation is ( dy dx )=p dy dx = ^ -1 (p) Integrating both sides with respect to x, we get: y = x ^ -1 (p) + C Given y(0) = q, substituting x = 0 and y = q: q = 0 + C C = q Substituting the value of C in the equation: y = x ^ -1 (p) + q y - q = x ^ -1 (p) y - q x = ^ -1 (p) ( y - q x ) = p Answer: ( y-q x )=p

Question 3

Passage: Consider the differential equation e^ x+y dy dx =e^ x-y . Question: What is the solution of the differential equation with y(0)=0 ?

Accepted Answer

The given differential equation is e^ x+y dy dx =e^ x-y e^x e^y dy dx = e^x e^ -y e^y dy dx = e^ -y e^ 2y dy = dx Integrating both sides: e^ 2y dy = dx e^ 2y 2 = x + C Given y(0)=0, substituting x=0 and y=0: e^0 2 = 0 + C C = 1 2 Substituting C back into the equation: e^ 2y 2 = x + 1 2 e^ 2y = 2x + 1 2y = (2x+1)

Question 4

What is the degree of the differential equation ( d^2y dx^2 )^ 3 2 = ( dy dx )^ 5 2 ?

Accepted Answer

The given differential equation is ( d^2y dx^2 )^ 3 2 = ( dy dx )^ 5 2 . To find the degree of the differential equation, the powers of the derivatives must be made integers. Squaring both sides, we get: ( d^2y dx^2 )^3 = ( dy dx )^5 The highest order derivative is d^2y dx^2 , which has an order of 2. The power of the highest order derivative is 3. Therefore, the degree of the differential equation is 3.

Question 5

If k is an arbitrary constant, then what is the general solution of the equation (x + y)^2 dy dx = k^2?

Accepted Answer

Let x + y = v Differentiating with respect to x, we get 1 + dy dx = dv dx dy dx = dv dx - 1 Substituting in the given differential equation: v^2 ( dv dx - 1 ) = k^2 v^2 dv dx = v^2 + k^2 v^2 v^2 + k^2 dv = dx ( 1 - k^2 v^2 + k^2 ) dv = dx Integrating both sides: ( 1 - k^2 v^2 + k^2 ) dv = dx v - k ^ -1 ( v k ) = x + c Substituting v = x + y: x + y - k ^ -1 ( x + y k ) = x + c y - c = k ^ -1 ( x + y k ) y - c k = ^ -1 ( x + y k ) x + y = k ( y - c k ) Answer: x + y = k ( y - c k )

Question 6

Consider the following statements: I. y = xe^ 2x is the solution of dy dx = y (2 + 1 x ). II. y = x |x| + cx is the solution of dy dx = x + y x . Which of the statements given above is/are correct?

Accepted Answer

For Statement I: Given y = xe^ 2x Differentiating with respect to x: dy dx = e^ 2x + 2xe^ 2x = e^ 2x (1 + 2x) Also, y (2 + 1 x ) = xe^ 2x ( 2x + 1 x ) = e^ 2x (2x + 1) Since dy dx = y (2 + 1 x ), Statement I is correct. For Statement II: Given y = x |x| + cx Differentiating with respect to x: dy dx = |x| + x ( 1 x ) + c = |x| + 1 + c Also, x + y x = 1 + y x = 1 + x |x| + cx x = 1 + |x| + c Since dy dx = x + y x , Statement II is correct. Both statements I and II are correct. Answer: Both I and I

Differential Equations — Previous Year Questions & Solutions