Let y1 and y2 be two different solutions of the equation dydx+P(x)y=Q(x). then αy1+βy2 Will be a solution of the given equation if 3(α+β)= ………………

ddx(αy1+βy2)+p(x)(αy1+βy2)=Q(x) ⇒α(dy1dx+p(x)y1)+β(dy2dx+p(x)y2)=Q(x) ⇒αQ(x)+βQ(x)=Q(x)⇒α+β=1

Let y1 and y2 be two different solutions of the equation dydx+P(x)y=Q(x). then αy1+βy2 Will be a solution of the given equation if 3(α+β)= ………………

ddx(αy1+βy2)+p(x)(αy1+βy2)=Q(x) ⇒α(dy1dx+p(x)y1)+β(dy2dx+p(x)y2)=Q(x) ⇒αQ(x)+βQ(x)=Q(x)⇒α+β=1

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Let $y_{1}$ and $y_{2}$ be two different solutions of the equation $\frac{d y}{d x} + P (x) y = Q (x) .$ then $α y_{1} + β y_{2}$ Will be a solution of the given equation if $3 (α + β) =$ ………………

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$\frac{d}{d x} (α y_{1} + β y_{2}) + p (x) (α y_{1} + β y_{2}) = Q (x) \Rightarrow α (\frac{d y_{1}}{d x} + p (x) y_{1}) + β (\frac{d y_{2}}{d x} + p (x) y_{2}) = Q (x) \Rightarrow α Q (x) + β Q (x) = Q (x) \Rightarrow α + β = 1$

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Let y1 and y2 be two different solutions of the equation dydx+P(x)y=Q(x). then αy1+βy2 Will be a solution of the given equation if 3(α+β)= ………………

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Let $y_{1}$ and $y_{2}$ be two different solutions of the equation $\frac{d y}{d x} + P (x) y = Q (x) .$ then $α y_{1} + β y_{2}$ Will be a solution of the given equation if $3 (α + β) =$ ………………