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In this lesson, we will learn how to solve initial value problems of differential equations.

Q1:

Find a particular solution for the following differential equation for which π¦ ( 0 ) = 1 2 :

Q2:

Find the solution for the following differential equation for π¦ ( 0 ) = 0 :

Q3:

Find the solution for the following differential equation for π¦ ( 1 ) = 1 :

Q4:

Find the solution for the following differential equation for π¦ ( 0 ) = ( 2 ) l n :

Q5:

Find the particular solution for the following separable differential equation: c o s d d ( π¦ ) π¦ π₯ β π₯ = 0 , π¦ ( 0 ) = 0 .

Q6:

Find the solution for the following differential equation for π¦ ( 0 ) = 1 6 :

Q7:

Find the solution for the following differential equation for π¦ ( 0 ) = 2 :

Q8:

Find the solution for the following differential equation for π¦ ( 0 ) = 1 :

Q9:

For a circuit containing a capacitor and a resistor, the first-order differential equation that describes a discharging capacitor is If π 0 represents the charge within the capacitor at π‘ = 0 s e c o n d s , find the general solution. πΆ is the capacitance of the capacitor. π is the resistance of the resistor.

Q10:

Suppose that d d s i n c o s π¦ π₯ = 5 2 π₯ ο» π¦ 3 ο 2 and π¦ = 3 π 4 when π₯ = π 6 . Find π¦ in terms of π₯ .

Q11:

Find the solution of the differential equation d d s e c π’ π‘ = π‘ + π‘ π’ 2 that satisfies the initial condition π’ ( 0 ) = β 3 .

Q12:

Find the solution of the differential equation π₯ π₯ = π¦ ο» 1 + β 1 + 3 π¦ ο π¦ β² l n 2 that satisfies the initial condition π¦ ( 1 ) = β 1 .

Q13:

Solve the differential equation d d π¦ π₯ οΉ π₯ + 4 ο = 3 ο¨ for π¦ given that π¦ ( 2 ) = 0 .

Q14:

The gradient of the tangent to a curve is ο β π¦ + 3 4 π₯ β 4 . Find the equation of the curve given that the curve passes through the point ( 5 , 3 ) .

Q15:

Find the solution of the differential equation π¦ β² π₯ = π + π¦ t a n , where 0 < π₯ < π 2 , that satisfies the initial condition π¦ ο» π 3 ο = π .

Q16:

Solve the differential equation d d π¦ π₯ β π₯ β 9 = 1 ο¨ for π¦ given that π¦ ( 5 ) = 3 l n .

Q17:

Find the solution of the differential equation d d l n πΏ π‘ = π πΏ π‘ 2 that satisfies the initial condition πΏ ( 1 ) = β 1 .

Q18:

Solve the differential equation π₯ π¦ π₯ = β π₯ β 4 d d ο¨ for π¦ given that π¦ ( 2 ) = 0 .

Q19:

Find the solution for the following differential equation for π¦ ( 2 ) = 1 :

Q20:

Find the solution of the differential equation π₯ + π¦ β π₯ + 3 π¦ π₯ = 0 2 2 d d that satisfies the initial condition π¦ ( β 1 ) = β 2 .

Q21:

Q22:

The gradient of the tangent of a curve is β 4 π₯ + 4 3 π¦ + 3 and the curve passes through the point ( β 2 , β 3 ) . Find the equation of the normal to the curve at the point where the π₯ -coordinate is β 2 .

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