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In this lesson, we will learn how to solve separable differential equations.

Q1:

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

Q2:

Solve the differential equation d d π¦ π₯ = β 5 π₯ π¦ 2 2 .

Q3:

Solve the differential equation d d l n π» π = π π» β 1 + π π» 2 2 .

Q4:

Find a relation between π¦ and π₯ , given that π₯ π¦ π¦ β² = π₯ β 5 2 .

Q5:

Find the equation of the curve that passes through the point ( β 3 , 2 ) given that the gradient of the tangent at any point is β 4 π₯ 7 π¦ .

Q6:

A relation π ( π₯ , π¦ ) = 0 is implicitly differentiated to obtain d d π¦ π₯ = 2 π₯ + 5 2 π¦ + 5 . Find the relation given that when π¦ = 3 , π₯ = 3 .

Q7:

Solve the differential equation d d π¦ π₯ + π¦ = 1 .

Q8:

Find the equation of the curve that passes through the point ( 0 , β 1 ) given d d π¦ π₯ = β 6 π₯ β 4 4 π¦ + 1 3 .

Q9:

Suppose that d d c o s s i n π¦ π₯ = 4 π₯ β 4 2 π₯ 4 π¦ + 9 and π¦ = 0 when π₯ = 0 . Find π¦ in terms of π₯ .

Q10:

Solve the following differential equation: ( π β 5 ) π¦ β² = 2 + π₯ π¦ c o s .

Q11:

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

Q12:

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

Q13:

Find the solution of the differential equation d d π¦ π₯ = π₯ π π¦ that satisfies the initial condition π¦ ( 0 ) = 0 .

Q14:

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

Q15:

Solve the following differential equation: d d π π‘ = π‘ π β 5 π + π‘ β 5 2 2 .

Q16:

Solve the differential equation d d s e c π π‘ = π‘ π π π π‘ 2 .

Q17:

Find the equation of the curve that passes through the point ( β 8 , 1 ) given that the gradient of the tangent at any point is equal to 2 times the square of the π¦ coordinate.

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