General
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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

Maths-General

  1. left parenthesis a plus m right parenthesis y equals e to the power of m x end exponent plus c e to the power of negative a x end exponent
  2. left parenthesis a plus m right parenthesis y equals e to the power of m x end exponent plus c
  3. y equals e to the power of m x end exponent plus c e to the power of negative a x end exponent
  4. y e to the power of a x end exponent equals m e to the power of m x end exponent plus c

    Answer:The correct answer is: left parenthesis a plus m right parenthesis y equals e to the power of m x end exponent plus c e to the power of negative a x end exponent

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    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
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    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
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    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

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    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
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    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

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

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

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    If a cooling curve is drawn taking t(s) along X-axis and temperature (q) along Y-axis, it appears as

    If a cooling curve is drawn taking t(s) along X-axis and temperature (q) along Y-axis, it appears as

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    Solution of differential equation xdy – ydx = 0 represents

    Solution of differential equation xdy – ydx = 0 represents

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    Two parallel plates of area A are separated by two different dielectric as shown in figure. The net capacitance is

    C equals fraction numerator epsilon subscript 0 end subscript A over denominator fraction numerator d subscript 1 end subscript over denominator K subscript 1 end subscript end fraction plus fraction numerator d subscript 2 end subscript over denominator K subscript 2 end subscript end fraction end fraction
    equals blank fraction numerator epsilon subscript 0 end subscript A over denominator fraction numerator d over denominator 2 end fraction open parentheses fraction numerator 1 over denominator 1 end fraction plus fraction numerator 1 over denominator 2 end fraction close parentheses end fraction equals fraction numerator 4 epsilon subscript 0 end subscript A over denominator 3 d end fraction

    Two parallel plates of area A are separated by two different dielectric as shown in figure. The net capacitance is

    physics-General
    C equals fraction numerator epsilon subscript 0 end subscript A over denominator fraction numerator d subscript 1 end subscript over denominator K subscript 1 end subscript end fraction plus fraction numerator d subscript 2 end subscript over denominator K subscript 2 end subscript end fraction end fraction
    equals blank fraction numerator epsilon subscript 0 end subscript A over denominator fraction numerator d over denominator 2 end fraction open parentheses fraction numerator 1 over denominator 1 end fraction plus fraction numerator 1 over denominator 2 end fraction close parentheses end fraction equals fraction numerator 4 epsilon subscript 0 end subscript A over denominator 3 d end fraction
    General
    physics-

    A network of six identical capacitors, each of valueC, is made as shown in the figure.
    The equivalent capacitance between the points A space a n d blank B is

    In the given circuit capacitor’s (1) (2) and (3) are connected in series, hence equivalent capacitance is

    fraction numerator 1 over denominator C to the power of ´ ´ end exponent end fraction equals fraction numerator 1 over denominator C end fraction plus fraction numerator 1 over denominator C end fraction plus fraction numerator 1 over denominator C end fraction equals fraction numerator 3 over denominator C end fraction
    rightwards double arrow C to the power of ´ ´ end exponent equals fraction numerator C over denominator 3 end fraction
    This is connected in parallel with (4).
    therefore blank C to the power of ´ ´ ´ ´ end exponent equals C to the power of ´ ´ end exponent plus C equals fraction numerator C over denominator 3 end fraction plus C equals fraction numerator 4 C over denominator 3 end fraction
    The three capacitor’s (5),fraction numerator 4 C over denominator 3 end fraction comma (6) are now connected in series.
    therefore Equivalent capacitance is
    fraction numerator 1 over denominator C blank ´ ´ ´ ´ ´ ´ end fraction equals fraction numerator 1 over denominator C end fraction plus fraction numerator 3 over denominator 4 C end fraction plus fraction numerator 1 over denominator C end fraction
    fraction numerator 1 over denominator C ´ ´ ´ ´ ´ ´ end fraction equals fraction numerator 11 over denominator 4 C end fraction
    rightwards double arrow blank C to the power of ´ ´ ´ ´ ´ ´ end exponent equals fraction numerator 4 C over denominator 11 end fraction

    A network of six identical capacitors, each of valueC, is made as shown in the figure.
    The equivalent capacitance between the points A space a n d blank B is

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    In the given circuit capacitor’s (1) (2) and (3) are connected in series, hence equivalent capacitance is

    fraction numerator 1 over denominator C to the power of ´ ´ end exponent end fraction equals fraction numerator 1 over denominator C end fraction plus fraction numerator 1 over denominator C end fraction plus fraction numerator 1 over denominator C end fraction equals fraction numerator 3 over denominator C end fraction
    rightwards double arrow C to the power of ´ ´ end exponent equals fraction numerator C over denominator 3 end fraction
    This is connected in parallel with (4).
    therefore blank C to the power of ´ ´ ´ ´ end exponent equals C to the power of ´ ´ end exponent plus C equals fraction numerator C over denominator 3 end fraction plus C equals fraction numerator 4 C over denominator 3 end fraction
    The three capacitor’s (5),fraction numerator 4 C over denominator 3 end fraction comma (6) are now connected in series.
    therefore Equivalent capacitance is
    fraction numerator 1 over denominator C blank ´ ´ ´ ´ ´ ´ end fraction equals fraction numerator 1 over denominator C end fraction plus fraction numerator 3 over denominator 4 C end fraction plus fraction numerator 1 over denominator C end fraction
    fraction numerator 1 over denominator C ´ ´ ´ ´ ´ ´ end fraction equals fraction numerator 11 over denominator 4 C end fraction
    rightwards double arrow blank C to the power of ´ ´ ´ ´ ´ ´ end exponent equals fraction numerator 4 C over denominator 11 end fraction
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    The equivalent capacitance between the points A a n d B in the following circuit is

    The two capacitors each of value 1.5mu F are in parallel. So, their equivalent capacitance

    Now, three capacitors each of value 3 mu F are in series. Hence, their equivalent capacitance is given by
    fraction numerator 1 over denominator C end fraction equals fraction numerator 1 over denominator 3 end fraction plus fraction numerator 1 over denominator 3 end fraction plus fraction numerator 1 over denominator 3 end fraction
    or fraction numerator 1 over denominator C end fraction blank equals fraction numerator 3 over denominator 3 end fraction
    or c equals 1 mu F

    The equivalent capacitance between the points A a n d B in the following circuit is

    physics-General
    The two capacitors each of value 1.5mu F are in parallel. So, their equivalent capacitance

    Now, three capacitors each of value 3 mu F are in series. Hence, their equivalent capacitance is given by
    fraction numerator 1 over denominator C end fraction equals fraction numerator 1 over denominator 3 end fraction plus fraction numerator 1 over denominator 3 end fraction plus fraction numerator 1 over denominator 3 end fraction
    or fraction numerator 1 over denominator C end fraction blank equals fraction numerator 3 over denominator 3 end fraction
    or c equals 1 mu F
    General
    physics-

    The equivalent capacitance of the combination shown in figure below is

    The three capacitors are in parallel hence, their equivalent capacitance equals 3 C

    The equivalent capacitance of the combination shown in figure below is

    physics-General
    The three capacitors are in parallel hence, their equivalent capacitance equals 3 C
    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-

    A capacitor of capacitance 1 blank mu F is filled with two dielectrics of dielectric constant 4 and 6. What is the new capacitance?

    Initially, the capacitance of capacitor

    C equals fraction numerator epsilon subscript 0 end subscript A over denominator d end fraction
    therefore blank fraction numerator epsilon subscript 0 end subscript A over denominator d end fraction equals 1 mu F …(i)
    When it is filled with dielectric of dielectric constant K subscript 1 end subscript a n d blank K subscript 2 end subscript as shown, then there are two capacitors connected is parallel. So,
    C to the power of ´ ´ end exponent equals fraction numerator K subscript 1 end subscript epsilon subscript 0 end subscript open parentheses fraction numerator A over denominator 2 end fraction close parentheses over denominator d end fraction plus fraction numerator K subscript 2 end subscript epsilon subscript 0 end subscript open parentheses fraction numerator A over denominator 2 end fraction close parentheses over denominator d end fraction
    (as area becomes half)
    C ´ ´ blank equals fraction numerator 4 epsilon subscript 0 end subscript A over denominator 2 d end fraction plus fraction numerator 6 epsilon subscript 0 end subscript A over denominator 2 d end fraction equals fraction numerator 2 epsilon subscript 0 end subscript A over denominator d end fraction plus fraction numerator 3 epsilon subscript 0 end subscript A over denominator d end fraction
    Using Eq. (i), we obtain
    C to the power of ´ ´ end exponent equals blank 2 cross times 1 plus 3 cross times 1 equals 5 blank mu F

    A capacitor of capacitance 1 blank mu F is filled with two dielectrics of dielectric constant 4 and 6. What is the new capacitance?

    physics-General
    Initially, the capacitance of capacitor

    C equals fraction numerator epsilon subscript 0 end subscript A over denominator d end fraction
    therefore blank fraction numerator epsilon subscript 0 end subscript A over denominator d end fraction equals 1 mu F …(i)
    When it is filled with dielectric of dielectric constant K subscript 1 end subscript a n d blank K subscript 2 end subscript as shown, then there are two capacitors connected is parallel. So,
    C to the power of ´ ´ end exponent equals fraction numerator K subscript 1 end subscript epsilon subscript 0 end subscript open parentheses fraction numerator A over denominator 2 end fraction close parentheses over denominator d end fraction plus fraction numerator K subscript 2 end subscript epsilon subscript 0 end subscript open parentheses fraction numerator A over denominator 2 end fraction close parentheses over denominator d end fraction
    (as area becomes half)
    C ´ ´ blank equals fraction numerator 4 epsilon subscript 0 end subscript A over denominator 2 d end fraction plus fraction numerator 6 epsilon subscript 0 end subscript A over denominator 2 d end fraction equals fraction numerator 2 epsilon subscript 0 end subscript A over denominator d end fraction plus fraction numerator 3 epsilon subscript 0 end subscript A over denominator d end fraction
    Using Eq. (i), we obtain
    C to the power of ´ ´ end exponent equals blank 2 cross times 1 plus 3 cross times 1 equals 5 blank mu F
    General
    physics-

    In the given network, the value of C, so that an equivalent capacitance between A and B i s blank 3 mu F comma is

    3 equals fraction numerator fraction numerator 16 over denominator 5 end fraction C over denominator fraction numerator 16 over denominator 5 end fraction plus C end fraction
    Or C equals 48 mu F

    In the given network, the value of C, so that an equivalent capacitance between A and B i s blank 3 mu F comma is

    physics-General
    3 equals fraction numerator fraction numerator 16 over denominator 5 end fraction C over denominator fraction numerator 16 over denominator 5 end fraction plus C end fraction
    Or C equals 48 mu F