Miscellaneous Exercise — Differential Equations

1. 13 marksNCERT Exercise

   For each of the differential equations given below, indicate its order and degree (if defined): (i) d^2y/dx^2+5x(dy/dx)^2-6y=log x (ii) (dy/dx)^3-4(dy/dx)^2+7y=sin x (iii) d^4y/dx^4-sin(d^3y/dx^3)=0

   Hint. Identify the highest derivative in each, then check whether each equation is a polynomial in that derivative.

   (i) The highest derivative is d^2y/dx^2 (order 2), raised to power 1, so degree 1. (ii) The highest derivative is dy/dx (order 1), raised to power 3, so degree 3. (iii) The highest derivative is d^4y/dx^4 (order 4), but since the equation involves sin of a lower derivative, it is not a polynomial equation in its derivatives, so the degree is not defined.

   ✦ (i) Order 2, degree 1. (ii) Order 1, degree 3. (iii) Order 4, degree not defined

2. 23 marksNCERT Exercise

   For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation: (i) xy=ae^x+be^-x+x^2 : x(d^2y/dx^2)+2(dy/dx)-xy+x^2-2=0 (ii) y=e^x(a cos x+b sin x) : d^2y/dx^2-2(dy/dx)+2y=0 (iii) y=x sin3x : d^2y/dx^2+9y-6cos3x=0 (iv) x^2=2y^2 log y : (x^2+y^2)(dy/dx)-xy=0

   Hint. Differentiate each given function the required number of times, then substitute into its corresponding differential equation and confirm both sides match.

   Since differentiating each given function (once or twice, as required) and substituting the resulting expressions for y, y', and y'' into its stated differential equation makes the left-hand side collapse to exactly zero, this confirms the given function is a solution in every one of the four cases.

   ✦ Verified in all four cases: (i), (ii), (iii), (iv) all satisfy their stated differential equations

3. 33 marksNCERT Exercise

   Prove that x^2-y^2=c(x^2+y^2)^2 is the general solution of the differential equation (x^3-3xy^2)dx=(y^3-3x^2y)dy, where c is a parameter.

   Hint. Differentiate the given relation implicitly with respect to x, substitute c back in terms of x and y, and simplify to recover the stated differential equation.

   Differentiating x^2-y^2=c(x^2+y^2)^2 implicitly and substituting c=(x^2-y^2)/(x^2+y^2)^2 back into the result, then clearing denominators and simplifying, gives exactly y'(y^3-3x^2y)=x^3-3xy^2, which is the given differential equation rearranged.

   ✦ Proved: the given relation satisfies (x^3-3xy^2)dx=(y^3-3x^2y)dy exactly

4. 43 marksNCERT Exercise

   Find the general solution of the differential equation dy/dx+sqrt[(1-y^2)/(1-x^2)]=0.

   Hint. Separate the variables — both sides reduce to a standard sin^-1 integral.

   Separating gives dy/sqrt(1-y^2)=-dx/sqrt(1-x^2), and integrating both sides using the standard sin^-1 form gives sin^-1y=-sin^-1x+C.

   ✦ sin^-1 x+sin^-1 y=C

5. 53 marksNCERT Exercise

   Show that the general solution of the differential equation dy/dx+(y^2+y+1)/(x^2+x+1)=0 is given by (x+y+1)=A(1-x-y-2xy), where A is a parameter.

   Hint. Separate the variables — both sides have the identical quadratic-plus-one form, integrating to matching tan^-1 expressions, then combine using the tan^-1 addition formula.

   Separating gives dy/(y^2+y+1)=-dx/(x^2+x+1), and since both sides share the same completed-square form, integrating gives tan^-1[(2y+1)/sqrt3]+tan^-1[(2x+1)/sqrt3]=C; applying the tan addition formula to combine the two inverse-tangent terms and simplifying the resulting algebraic relation yields exactly the stated form.

   ✦ Proved: (x+y+1)=A(1-x-y-2xy)

6. 63 marksNCERT Exercise

   Find the equation of the curve passing through the point (0,pi/4) whose differential equation is sin x cos y dx+cos x sin y dy=0.

   Hint. Separate the variables — divide through by cos x cos y, reducing both sides to a tan integral.

   Dividing by cosx cosy gives tanx dx+tany dy=0, and integrating both sides gives log|secx|+log|secy|=C1, which rearranges to cosx cosy=C; applying the point (0,pi/4) fixes the constant.

   ✦ cos x . cos y=sqrt(2)/2

7. 73 marksNCERT Exercise

   Find the particular solution of the differential equation (1+e^(2x))dy+(1+y^2)e^x dx=0, given that y=1 when x=0.

   Hint. Separate the variables — the x-side integral uses the substitution u=e^x to reduce to a standard tan^-1 form.

   Separating gives dy/(1+y^2)=-e^x dx/(1+e^(2x)), and integrating both sides (the right via u=e^x) gives tan^-1y=-tan^-1(e^x)+C; applying y(0)=1 gives C=pi/2, and simplifying tan^-1y=pi/2-tan^-1(e^x) using the complementary-angle identity gives a strikingly clean explicit solution.

   ✦ y=e^(-x)

8. 83 marksNCERT Exercise

   Solve the differential equation y.e^(x/y) dx=(x.e^(x/y)+y^2)dy (y is not 0).

   Hint. This is homogeneous in x as a function of y — put x=vy, which reduces the equation to a directly separable one in e^v.

   With x=vy, the equation reduces to y(dv/dy)=e^(-v), a separable equation whose left side (e^v dv) integrates directly to e^v; substituting v=x/y back gives the general solution.

   ✦ e^(x/y)=y+C

9. 93 marksNCERT Exercise

   Find a particular solution of the differential equation (x-y)(dx+dy)=dx-dy, given that y=-1 when x=0. [Hint: put x-y=t]

   Hint. Following the hint, put t=x-y, express dx+dy in terms of dx and dt, and reduce to a separable equation in t and x.

   With t=x-y, dy=dx-dt, so dx+dy=2dx-dt; substituting into t(dx+dy)=dt gives dx=[1/(2t)+1/2]dt, which integrates to x=(1/2)log|t|+t/2+C; applying the point (0,-1) (where t=1) fixes the constant.

   ✦ log|x-y|=x+y+1

10. 103 marksNCERT Exercise

    Solve the differential equation [e^(-2sqrt(x))/sqrt(x) - y/sqrt(x)] dx/dy=1 (x is not 0).

    Hint. Invert to write this as a linear equation in y(x): dy/dx+y/sqrt(x)=e^(-2sqrt(x))/sqrt(x), with integrating factor e^(2sqrt(x)).

    Rewriting gives dy/dx+y/sqrt(x)=e^(-2sqrt(x))/sqrt(x), linear with integrating factor e^(integral of dx/sqrt(x))=e^(2sqrt(x)); integrating y.e^(2sqrt(x))=integral of dx/sqrt(x)=2sqrt(x)+C gives the general solution.

    ✦ y=(2sqrt(x)+C)e^(-2sqrt(x))

11. 113 marksNCERT Exercise

    Find a particular solution of the differential equation dy/dx+y cot x=4x cosec x (x is not 0), given that y=0 when x=pi/2.

    Hint. This is linear with integrating factor sinx; integrate then apply the given condition.

    The integrating factor is e^(integral of cotx dx)=sinx; integrating y sinx=integral of 4x dx=2x^2+C, then applying y(pi/2)=0, fixes the constant.

    ✦ y=(2x^2-pi^2/2)/sin x, i.e. y sin x=2x^2-pi^2/2

12. 123 marksNCERT Exercise

    Find a particular solution of the differential equation (x+1)dy/dx=2e^(-y)-1, given that y=0 when x=0.

    Hint. Separate the variables — divide through by (2e^-y-1) and by (x+1).

    Separating gives e^y dy/(2-e^y)=dx/(x+1), and integrating both sides (the left via t=2-e^y) gives -log|2-e^y|=log|x+1|+C1; applying y(0)=0 and simplifying gives an explicit form for y.

    ✦ y=log[2-1/(x+1)]

13. 133 marksNCERT Exercise

    Choose the correct answer: the general solution of the differential equation (y dx-x dy)/y=0 is (A) xy=C (B) x=Cy^2 (C) y=Cx (D) y=Cx^2

    Hint. Separate the variables directly — this reduces to dx/x=dy/y.

    Since y dx=x dy, dividing both sides by xy gives dx/x=dy/y, and integrating gives log x=log y+C1, which exponentiates to x=Cy, equivalently y=Cx after relabelling the constant.

    ✦ (C) y=Cx

14. 143 marksNCERT Exercise

    Choose the correct answer: the general solution of a differential equation of the type dx/dy+P1x=Q1 is (A) y.e^(integral P1 dx)=integral(Q1 e^(integral P1 dx))dx+C (B) y.e^(integral P1 dx)=integral(Q1 e^(integral P1 dx))dx+C (C) x.e^(integral P1 dy)=integral(Q1 e^(integral P1 dy))dy+C (D) x.e^(integral P1 dx)=integral(Q1 e^(integral P1 dx))dx+C

    Hint. Recall that when the equation is linear in x as a function of y, the integrating factor and the general-solution formula are both built from integrals with respect to y, not x.

    Since dx/dy+P1x=Q1 treats x as the dependent variable and y as the independent variable, both the integrating factor e^(integral of P1 dy) and the final integral on the right must be taken with respect to y, matching exactly one option.

    ✦ (C) x.e^(integral P1 dy)=integral(Q1.e^(integral P1 dy))dy+C

15. 153 marksNCERT Exercise

    Choose the correct answer: the general solution of the differential equation e^x dy+(y.e^x+2x)dx=0 is (A) x.e^y+x^2=C (B) x.e^y+y^2=C (C) y.e^x+x^2=C (D) y.e^y+x^2=C

    Hint. Divide through by e^x first to put the equation into standard linear form dy/dx+y=-2x.e^-x, then find the integrating factor.

    Dividing by e^x gives dy/dx+y=-2x.e^-x, linear with integrating factor e^x; integrating y.e^x=integral of -2x dx=-x^2+C gives y.e^x+x^2=C directly.

    ✦ (C) y.e^x+x^2=C

Solutions written by the tuition.in editorial team and checked against lemh203.pdf (NCERT, Reprint 2026-27). Questions are referenced from the NCERT textbook for identification.