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Maths-

Solution of differential equation d y minus sin space x sin space x d x equals 0 is

Maths-General

  1. e to the power of cos space x end exponent tan space y equals c
  2. e to the power of cos space x end exponent tan space y over 2 equals c
  3. cos space x sin space y equals c
  4. cos space x tan space y equals c

    Answer:The correct answer is: e to the power of cos space x end exponent tan space y over 2 equals c

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    Related Questions to study

    General
    physics-

    In given circuit when switch S has been closed then charge on capacitor A and B respectively are

    The circuit is given as

    Let q subscript 1 end subscript a n d q subscript 2 end subscript be the charge after switch S has been closed.
    Then, V equals fraction numerator q subscript 1 end subscript over denominator 6 C end fraction equals fraction numerator q subscript 2 end subscript over denominator 3 C end fraction
    rightwards double arrow blank fraction numerator q subscript 1 end subscript over denominator 2 end fraction equals q subscript 2 end subscript
    rightwards double arrow blank q subscript 1 end subscript equals 2 q subscript 2 end subscript…(i)
    But we know that, charge is conserved
    q subscript 1 end subscript plus q subscript 2 end subscript equals 3 q plus 6 q
    or q subscript 1 end subscript plus q subscript 2 end subscript equals 9 q…(ii)
    On putting the value of q subscript 1 end subscriptEq. (ii)
    2 q subscript 2 end subscript plus q subscript 2 end subscript equals 9 q
    rightwards double arrow 3 q subscript 2 end subscript equals 9 q
    q subscript 2 end subscript equals 3 q
    Now, from Eq. (i)
    q subscript 1 end subscript equals 2 cross times 3 q
    rightwards double arrow q subscript 1 end subscript equals 6 q
    Hence,q subscript 1 end subscript equals 6 q comma blank q subscript 2 end subscript equals 3 q

    In given circuit when switch S has been closed then charge on capacitor A and B respectively are

    physics-General
    The circuit is given as

    Let q subscript 1 end subscript a n d q subscript 2 end subscript be the charge after switch S has been closed.
    Then, V equals fraction numerator q subscript 1 end subscript over denominator 6 C end fraction equals fraction numerator q subscript 2 end subscript over denominator 3 C end fraction
    rightwards double arrow blank fraction numerator q subscript 1 end subscript over denominator 2 end fraction equals q subscript 2 end subscript
    rightwards double arrow blank q subscript 1 end subscript equals 2 q subscript 2 end subscript…(i)
    But we know that, charge is conserved
    q subscript 1 end subscript plus q subscript 2 end subscript equals 3 q plus 6 q
    or q subscript 1 end subscript plus q subscript 2 end subscript equals 9 q…(ii)
    On putting the value of q subscript 1 end subscriptEq. (ii)
    2 q subscript 2 end subscript plus q subscript 2 end subscript equals 9 q
    rightwards double arrow 3 q subscript 2 end subscript equals 9 q
    q subscript 2 end subscript equals 3 q
    Now, from Eq. (i)
    q subscript 1 end subscript equals 2 cross times 3 q
    rightwards double arrow q subscript 1 end subscript equals 6 q
    Hence,q subscript 1 end subscript equals 6 q comma blank q subscript 2 end subscript equals 3 q
    General
    physics-

    In the given figure, a hollow spherical capacitor is shown. The electric field will not be zero at

    The electric field of a hollow spherical capacitor is localised in between inner and outer surface of the spherical conductor.
    Therefore, at point r subscript 1 end subscript less than r less than r subscript 2 end subscript, the electric field will not be zero.

    In the given figure, a hollow spherical capacitor is shown. The electric field will not be zero at

    physics-General
    The electric field of a hollow spherical capacitor is localised in between inner and outer surface of the spherical conductor.
    Therefore, at point r subscript 1 end subscript less than r less than r subscript 2 end subscript, the electric field will not be zero.
    General
    physics-

    In the capacitor shown in the circuit is changed to 5 V and left in the circuit, in 12s the charge on the capacitor will become left parenthesis e equals 2.718 right parenthesis

    Final charge on capacitor is
    q equals q subscript 0 end subscript e to the power of negative t divided by R C end exponent
    where q subscript 0 end subscript= charge on capacitor at t equals 0 comma
    R C=time constant of the circuit.
    Putting q subscript 0 end subscript equals C V subscript 0 end subscript
    therefore q equals C V subscript 0 end subscript superscript negative 1 divided by R C end superscript
    Given, C equals 2 F comma V subscript 0 end subscript equals 5 blank v o l t comma R equals 6 blank capital omega comma blank t equals 12 blank s
    Hence, q equals open parentheses 2 cross times 5 close parentheses e to the power of negative left parenthesis 12 divided by 6 cross times 2 right parenthesis end exponent
    equals 10 e to the power of negative 1 end exponent equals fraction numerator 10 over denominator e end fraction C

    In the capacitor shown in the circuit is changed to 5 V and left in the circuit, in 12s the charge on the capacitor will become left parenthesis e equals 2.718 right parenthesis

    physics-General
    Final charge on capacitor is
    q equals q subscript 0 end subscript e to the power of negative t divided by R C end exponent
    where q subscript 0 end subscript= charge on capacitor at t equals 0 comma
    R C=time constant of the circuit.
    Putting q subscript 0 end subscript equals C V subscript 0 end subscript
    therefore q equals C V subscript 0 end subscript superscript negative 1 divided by R C end superscript
    Given, C equals 2 F comma V subscript 0 end subscript equals 5 blank v o l t comma R equals 6 blank capital omega comma blank t equals 12 blank s
    Hence, q equals open parentheses 2 cross times 5 close parentheses e to the power of negative left parenthesis 12 divided by 6 cross times 2 right parenthesis end exponent
    equals 10 e to the power of negative 1 end exponent equals fraction numerator 10 over denominator e end fraction C
    General
    maths-

    The solution of the differential equation 2 x fraction numerator d y over denominator d x end fraction minus y equals 3 represent

    The solution of the differential equation 2 x fraction numerator d y over denominator d x end fraction minus y equals 3 represent

    maths-General
    General
    maths-

    Solution of differential equation fraction numerator d y over denominator d x end fraction plus a y equals e to the power of m x end exponent is

    Solution of differential equation fraction numerator d y over denominator d x end fraction plus a y equals e to the power of m x end exponent is

    maths-General
    General
    physics-

    The stress - strain graphs for materials A and B are as shown. Choose the correct alternative

    The stress - strain graphs for materials A and B are as shown. Choose the correct alternative

    physics-General
    General
    physics-

    In the experiment to determine Young’s modulus of the material of a wire under tension used in the arrangement as shown. The percentage error in the measurement of length is ‘a’, in the measurement of the radius of the wire is ‘b’ and in the measurement of the change in length of the wire is ‘c’. Percentage error in the measurement of Young’s modulus for a given load is

    In the experiment to determine Young’s modulus of the material of a wire under tension used in the arrangement as shown. The percentage error in the measurement of length is ‘a’, in the measurement of the radius of the wire is ‘b’ and in the measurement of the change in length of the wire is ‘c’. Percentage error in the measurement of Young’s modulus for a given load is

    physics-General
    General
    physics-

    The graph shows the change ' ' Dl in the length of a thin uniform wire used by the application of force ‘F’ at different temperatures T1 and T2 The variation suggests that

    The graph shows the change ' ' Dl in the length of a thin uniform wire used by the application of force ‘F’ at different temperatures T1 and T2 The variation suggests that

    physics-General
    General
    physics-

    The load versus extension graph for four wires of same material is shown. The thinnest wire is represented by the line

    The load versus extension graph for four wires of same material is shown. The thinnest wire is represented by the line

    physics-General
    General
    physics-

    Four plates of equal area A are separated by equal distance d and are arranged as shown in the figure. The equivalent capacity is

    The given circuit is equivalent to a parallel combination of two identical capacitors.

    Hence, equivalent capacitance between points A a n d blank B is
    C equals fraction numerator epsilon subscript 0 end subscript A over denominator d end fraction plus fraction numerator epsilon subscript 0 end subscript A over denominator d end fraction equals fraction numerator 2 epsilon subscript 0 end subscript A over denominator d end fraction

    Four plates of equal area A are separated by equal distance d and are arranged as shown in the figure. The equivalent capacity is

    physics-General
    The given circuit is equivalent to a parallel combination of two identical capacitors.

    Hence, equivalent capacitance between points A a n d blank B is
    C equals fraction numerator epsilon subscript 0 end subscript A over denominator d end fraction plus fraction numerator epsilon subscript 0 end subscript A over denominator d end fraction equals fraction numerator 2 epsilon subscript 0 end subscript A over denominator d end fraction
    General
    physics-

    The equivalent capacitance of the combination of the capacitors is

    The 10mu F and 6mu F capacitors are connected in parallel, hence resultant capacitance is
    C to the power of ´ end exponent equals 10 blank mu F plus 6 blank mu F equals 16 blank mu F blank
    This is connected in series with 4mu F blankcapacitor, hence effective capacitance is
    fraction numerator 1 over denominator C ´ ´ end fraction equals fraction numerator 1 over denominator 16 end fraction plus fraction numerator 1 over denominator 4 end fraction equals fraction numerator 20 over denominator 16 cross times 4 end fraction
    rightwards double arrow blank C to the power of ´ ´ end exponent equals fraction numerator 64 over denominator 20 end fraction equals 3.20 mu F

    The equivalent capacitance of the combination of the capacitors is

    physics-General
    The 10mu F and 6mu F capacitors are connected in parallel, hence resultant capacitance is
    C to the power of ´ end exponent equals 10 blank mu F plus 6 blank mu F equals 16 blank mu F blank
    This is connected in series with 4mu F blankcapacitor, hence effective capacitance is
    fraction numerator 1 over denominator C ´ ´ end fraction equals fraction numerator 1 over denominator 16 end fraction plus fraction numerator 1 over denominator 4 end fraction equals fraction numerator 20 over denominator 16 cross times 4 end fraction
    rightwards double arrow blank C to the power of ´ ´ end exponent equals fraction numerator 64 over denominator 20 end fraction equals 3.20 mu F
    General
    maths-

    Integrating factor of the differential equation cos space x fraction numerator d y over denominator d x end fraction plus y sin space x equals 1 is

    Integrating factor of the differential equation cos space x fraction numerator d y over denominator d x end fraction plus y sin space x equals 1 is

    maths-General
    General
    physics-

    A 2mu F capacitor is charged as shown in the figure. The percentage of its stored energy dissipated after the switch s is turned to positions 2 is

    q subscript i end subscript equals C subscript i end subscript V equals 2 V equals q (say)
    This charge will remain constant after switch is shifted from position 1 to position 2.
    U subscript i end subscript equals fraction numerator 1 over denominator 2 end fraction fraction numerator q to the power of 2 end exponent over denominator C subscript i end subscript end fraction equals fraction numerator q to the power of 2 end exponent over denominator 2 cross times 2 end fraction equals fraction numerator q to the power of 2 end exponent over denominator 4 end fraction
    U subscript i end subscript equals fraction numerator 1 over denominator 2 end fraction fraction numerator q to the power of 2 end exponent over denominator C subscript i end subscript end fraction equals fraction numerator q to the power of 2 end exponent over denominator 2 cross times 10 end fraction equals fraction numerator q to the power of 2 end exponent over denominator 20 end fraction
    thereforeEnergy dissipatedopen parentheses fraction numerator q to the power of 2 end exponent over denominator 5 end fraction close parentheses i s blank80% of the initial stored energy open parentheses equals fraction numerator q to the power of 2 end exponent over denominator 4 end fraction close parentheses.

    A 2mu F capacitor is charged as shown in the figure. The percentage of its stored energy dissipated after the switch s is turned to positions 2 is

    physics-General
    q subscript i end subscript equals C subscript i end subscript V equals 2 V equals q (say)
    This charge will remain constant after switch is shifted from position 1 to position 2.
    U subscript i end subscript equals fraction numerator 1 over denominator 2 end fraction fraction numerator q to the power of 2 end exponent over denominator C subscript i end subscript end fraction equals fraction numerator q to the power of 2 end exponent over denominator 2 cross times 2 end fraction equals fraction numerator q to the power of 2 end exponent over denominator 4 end fraction
    U subscript i end subscript equals fraction numerator 1 over denominator 2 end fraction fraction numerator q to the power of 2 end exponent over denominator C subscript i end subscript end fraction equals fraction numerator q to the power of 2 end exponent over denominator 2 cross times 10 end fraction equals fraction numerator q to the power of 2 end exponent over denominator 20 end fraction
    thereforeEnergy dissipatedopen parentheses fraction numerator q to the power of 2 end exponent over denominator 5 end fraction close parentheses i s blank80% of the initial stored energy open parentheses equals fraction numerator q to the power of 2 end exponent over denominator 4 end fraction close parentheses.
    General
    physics-

    A lead shot of diameter 1mm falls through a long column of glycerine The variation of the velocity ‘v’ with distance covered (s) is represented by

    A lead shot of diameter 1mm falls through a long column of glycerine The variation of the velocity ‘v’ with distance covered (s) is represented by

    physics-General
    General
    physics-

    What is the potential difference between points A a n d B in the circuit shown?

    Consider the charge distribution as shown. Considering the branch on upper side, we have

    fraction numerator q over denominator V subscript x end subscript minus V subscript A end subscript end fraction equals 4 cross times 10 to the power of negative 6 end exponent
    fraction numerator q over denominator V subscript A end subscript minus V subscript y end subscript end fraction equals blank 2 cross times 10 to the power of negative 6 end exponent
    Here, V subscript x end subscript equals 6 blank v o l t comma blank V subscript y end subscript equals 0
    therefore fraction numerator q over denominator 6 minus V subscript A end subscript end fraction equals 4 cross times 10 to the power of negative 6 end exponent…(i)
    fraction numerator q over denominator V subscript A end subscript minus 0 end fraction equals 2 cross times 10 to the power of negative 6 end exponent …(ii)
    From Eqs. (i) and (ii), we get
    fraction numerator V subscript A end subscript over denominator 6 minus V subscript A end subscript end fraction equals 2
    therefore V subscript A end subscript equals 4 v o l t
    Similarly for the lower side branch
    fraction numerator q ´ ´ over denominator 6 minus V subscript B end subscript end fraction equals blank 2 cross times 10 to the power of negative 6 end exponent…(iii)
    fraction numerator q ´ ´ over denominator V subscript B end subscript minus 0 end fraction equals blank 4 cross times 10 to the power of negative 6 end exponent...(iv)
    From Eqs. (iii) and (iv)
    fraction numerator V subscript B end subscript over denominator 6 minus V subscript B end subscript end fraction equals fraction numerator 1 over denominator 2 end fraction
    therefore blank V subscript B end subscript equals 2 blank v o l t
    therefore V subscript A end subscript minus V subscript B end subscript equals 4 minus 2 equals 2 blank v o l t

    What is the potential difference between points A a n d B in the circuit shown?

    physics-General
    Consider the charge distribution as shown. Considering the branch on upper side, we have

    fraction numerator q over denominator V subscript x end subscript minus V subscript A end subscript end fraction equals 4 cross times 10 to the power of negative 6 end exponent
    fraction numerator q over denominator V subscript A end subscript minus V subscript y end subscript end fraction equals blank 2 cross times 10 to the power of negative 6 end exponent
    Here, V subscript x end subscript equals 6 blank v o l t comma blank V subscript y end subscript equals 0
    therefore fraction numerator q over denominator 6 minus V subscript A end subscript end fraction equals 4 cross times 10 to the power of negative 6 end exponent…(i)
    fraction numerator q over denominator V subscript A end subscript minus 0 end fraction equals 2 cross times 10 to the power of negative 6 end exponent …(ii)
    From Eqs. (i) and (ii), we get
    fraction numerator V subscript A end subscript over denominator 6 minus V subscript A end subscript end fraction equals 2
    therefore V subscript A end subscript equals 4 v o l t
    Similarly for the lower side branch
    fraction numerator q ´ ´ over denominator 6 minus V subscript B end subscript end fraction equals blank 2 cross times 10 to the power of negative 6 end exponent…(iii)
    fraction numerator q ´ ´ over denominator V subscript B end subscript minus 0 end fraction equals blank 4 cross times 10 to the power of negative 6 end exponent...(iv)
    From Eqs. (iii) and (iv)
    fraction numerator V subscript B end subscript over denominator 6 minus V subscript B end subscript end fraction equals fraction numerator 1 over denominator 2 end fraction
    therefore blank V subscript B end subscript equals 2 blank v o l t
    therefore V subscript A end subscript minus V subscript B end subscript equals 4 minus 2 equals 2 blank v o l t